My answer is going to focus on the energy requirements of your ship which touch on the relative amount of space you need for getting the ship to its destination and supplying power to your people.
Summary
Solar power is useless, so don't bother. Either antimatter or fusion reactors would be perfectly useful for personnel energy demands, and would consist of a very negligible amount of the overall size and mass of the ship.
Your 1 gram of antimatter per day will support at least 2 million people. It will also require total hand-waving to acquire that midflight.
However, any kind of nuclear fusion, and a paltry 1 gram of antimatter per day, will be completely worthless for powering the ship, assuming we're hitting top speed within 10 years. If you want it to take much longer time frames, it might be possible, but I didn't calculate that.
You'll need to bring about 10% of your ship's mass worth of antimatter to hit 0.1 c in 10 years, or about 350% of your ship's mass worth of antimatter to hit 0.9 c in 10 years. I don't think 0.99 c is remotely doable without extreme advances in propulsion technology.
That much antimatter would take a few billion times the current age of the universe to produce at current rates, so you'll need way better tech. Still, the Sun outputs plenty of energy to accomplish it if you can build enough generators.
Fusion Power
At this point, we're outside the range of hard science. We know generally how fusion works, but we've never done it in a lab in a sustainable form. (We've done fusion, but it takes more power than the fusion produces, so it's an awesome experiment, but utterly worthless as a power source.)
That said, this MIT experiment is estimated to produce a lot of power if they ever make it work.
A working ARC fusion reactor would use 50 megawatts (MW) of power to produce 500MW of fusion power, 200MW of which could be delivered to the grid. That's enough to provide 200,000 people with electricity.
The reactor itself is about 1 meter across, so we don't need to worry about its mass too much. The infrastructure for ITER's reactor is about three stories tall, but a few dozens rooms worth of space for every 200k people is negligible.
The Culham Center for Fusion Energy estimates
A large power station generating 1,500 megawatts of electricity would consume approximately 600 grammes of tritium and 400 grammes of deuterium each day.
That equates to about 0.243 kg of fuel per megawatt per year. Given the 1 MW per thousand people figure on the MIT article, that's 243 kg fuel per million people per year, which is pretty negligible.
Antimatter Power
As I pointed out in the answer to this other question of yours (and pointed out in John Dallman's comment above), creating antimatter to use as a power source doesn't really make any sense. The power used to create the antimatter is millions to billions of times higher than what you finally get out of the antimatter annihilation.
You could use some kind of hypothetical device that collects antimatter with total handwavium (say, there's enough antimatter hanging out in interstellar space that you can just grab it on the way by, or zero-point energy). In that case we can calculate the energy from 1 gram per day (about 50 gigawatt hours per day which converts to about 2 GW total output). But none of that is remotely hard science.
From the section on Fusion Power, humans living in modern Boston use about 1 MW per thousand people, 2 GW would provide for 2 million people. That number would likely be far lower in an actual generational ship as people would learn how to do more with less. Still, it's a good upper bound.
Importantly, one gram per year per 2 million people means the normal matter mass you'd need to annihilate the matter is negligible. Presumably, you'd have some kind of reactor that takes space and mass, but since we don't have antimatter collectors and/or generators it's hard to say exactly how much. I'll assume it's about the same as a fusion reactor.
Without any handwavium, you'd need to bring the antimatter with you. The amount of 0.5 g per year per 2 million people is going to be totally negligible in terms of size and mass, but will require some very advanced means of actually producing that much antimatter.
Solar Power
Also, as pointed out in John Dallman's comment, solar panels are probably a huge waste. At 93 million miles from the Sun, we're seeing about 1.3 kW per square meter. At 0.1 c over a single generation (about 28 years), you'd travel about 16 trillion miles, which happens to be about halfway to the nearest star. Power output will fall off with the square of distance, so you're looking at about 44 nanowatts per m² at that distance, and an average of 7.6 milliwatts per m² across the trip.
Even if you could somehow fly in a line that gets you really close to each star you pass, your best case is going to be about 1.6 watts per m², assuming you're literally touching the surface of each star on the way past.
To be fair, not all stars are like ours in power output, but the Sun is actually the top 10% by mass, so your realistic solar influx will be even lower than the calculations above. Further, your realistic path will probably stay substantially farther from nearby stars than the calculations above, further lowering the average power.
From Sunmetrix, typical solar panels are 10-20 kg per m². At the lower value, you're looking at about 6 million kg per megawatt) in the best case of zooming right up to sun-like stars.
In addition to the mass issues, you have to have some way to spread them over an enormous area without shearing or folding from the torque. One megawatt is 600 thousand square meters in our best case scenario. That fits into a circle with a 437 meter radius.
If the ship is accelerating at 0.01 gees, a 1 m² section at the edge, with its mass of 10 kg, requires about 1 N force to keep it in place. At 437 meters from the center, that's 437 N-m of torque per m². There are about $2\pi r$ of these 1 m² sections around the outer radius. Then $2\pi (r-1)$ sections around a slightly smaller section. Turning that into an integral gives us about 600k N-m torque on the center of the disc.
You could probably solve the torque issues for a 437 meter disc by using supporting structures and so forth. But you need a thousand such discs for each million people on your ship. And realistically, you're looking at something closer to the 44 nW per m² figure. That requires some 23 trillion m² of panels per MW, or 23 quadrillion m² of panels per million people. Which ends up with a 151k km radius array, which has about 16% of the area between Earth and the Moon's orbit. The total torque is about $72\cdot10^{15}N-m$ and you're really not getting around that with extra supports, unless your entire ship is about that large.
As a side note, solar panels have a best-case efficiency of about 86%, and are realistically sitting around 50%. Your advanced people could likely hit 70-80%, but this is pretty trivial when there's so little sunlight available in the first place.
Acceleration Energy Requirements
Ok, so we need a negligible amount of extra space for the fusion and antimatter reactors compared to personnel energy usage. But we still need to accelerate the ship.
To hit 0.1 c in ten years, we need about 0.1 g acceleration.
To get one megaton of mass to 0.1 c, we need about $4.49\cdot10^{23} J$. That requires about 5 million kg, or five kilotons, worth of energy.
For the anti-matter propulsion system, the extra mass for propulsion is pretty negligible at about 0.5%.
For nuclear fusion, we're getting about 1.5 GJ per kg of fuel. That means about $3\cdot10^{14}kg$ of fuel. Which means about 3 parts per million of the spaceship's mass is payload; the rest is fuel. So really, antimatter rockets are the only way we're getting this ship to 0.1 c.
If we up the cruise velocity to 0.9 c, we'll need 0.4 megatons worth of energy. That's huge, but doable, in the sense that your rocket will still be 71% payload.
On the other hand, getting 0.2 megatons of antimatter is insane. With current antimatter production methods that use 15 billion times the antimatter's mass energy to create it, at a rate of 1 billion years per gram, you'd need about 22 times the Sun's annual output of energy, and way more particle accelerators than we currently have to accomplish it before the Sun dies. That is extremely advanced technology, but it seems plausible to an advanced enough society.
From Wikipedia, chemical rockets have an energy efficiency of about 60%. Anti-matter rockets have between 10 and 85% efficiency. But it really doesn't matter; anti-matter will have negligible mass, fusion will be way too high.
Reaction Mass
Now, you'll need reaction mass dependent on how much energy you can impart into each particle, and is given by $M=P\left(e^{\frac{\Delta v}{v_e}}-1\right)$, where $M$ is reaction mass, $P$ is payload mass, $\Delta v$ is the change in spaceship velocity, and $v_e$ is the exhaust velocity.
We've set $\Delta v=0.1c$. From this page, antimatter rockets have a specific impulse of 0.6c. From what I can tell, they're using "specific impulse" to mean "effective exhaust velocity", so $v_e=0.6c$. Plugging this into the equation, we get 18% P. Calculating a final velocity of 0.9 c requires about 78% of the ship to be reaction mass.
If we use dense reaction mass, this means the actual size of the spaceship isn't largely affected by the reaction mass. And adding 20% mass to the ship isn't particularly substantial in the grand scheme of material requirements.
From here, fusion rockets have exhaust velocities up to 700 km/s, or about 0.0023 c. This brings our required reaction mass up a lot (about 7 billion billion P). No way we're doing anything with that.
So if your guys are using fusion reactors to power the spaceship, you'll have to assume they achieve much higher exhaust velocities. Around 0.14 c is required to keep the reaction mass lower than the payload mass. This seems reasonable enough for an advanced society, but we don't have any current means of achieving it. Not that fusion was ever a real choice considering how much mass we need to power it.
