Hoo boy. This is one serious orbital mechanics problem.
The most energy-efficient method of getting a spacecraft (or, in this case, an asteroid) from one roughly-circular orbit into another is with a Hohmann transfer. For moving Ceres into Mars's orbit, this will entail firing thrusters on Ceres directly opposing its direction of motion so that its perihelion (its closest approach to the Sun) just touches Mars's orbit, waiting until Ceres reaches that point, then firing the thrusters again to circularize the orbit.
In order to actually have Ceres fall into Mars's orbit, though, the maneuver must be initiated at exactly the right time, so that when Ceres completes its half-of-an-ellipse transfer orbit, Mars will be right there waiting for it. Ceres has an orbital period of 4.60 Earth years, while Mars's year is only 1.8808 Earth years. They line up about every Mars year and a half, and the transfer itself will take less than half of a Ceres year. If there were rockets planted on Ceres' surface right now, that means Ceres could be in orbit around Mars within 8 Earth-years in the worst-case scenario, where the most recent launch window just recently closed. There's plenty of time to prepare.
The most important quantity in orbital mechanics is delta-V, which basically just measures the amount by which your spacecraft (or asteroid) needs to change its speed, which, in turn, determines how much fuel you need, how much that fuel and the engines used to burn it will weigh, how much more fuel you need to move all that fuel around, etc. It's used a bit like how distances are used when traveling around the Earth.
That Wikipedia page gives the delta-V for the Hohmann transfer as follows:
$$\Delta v_1 = \sqrt{\frac \mu {r_1}} \left( \sqrt{\frac {2r_2} {r_1+r_2}} - 1 \right)$$
$$\Delta v_2 = \sqrt{\frac \mu {r_2}} \left( 1 - \sqrt{\frac {2r_1} {r_1+r_2}} \right)$$
where $\Delta v_1$ is the delta-V needed to put the asteroid into the transfer orbit, $\Delta v_2$ is the delta-V needed to synch that orbit up with Mars, $\mu$ is the mass of the Sun multiplied by the gravitational constant G, $r_1$ is the radius of Ceres' current orbit, and $r_2$ is the radius of Mars's orbit.
Plugging those equations into Wolfram|Alpha, we get $\Delta v_1$ = 2.814 km/s and $\Delta v_2$ = 3.272 km/s, for a total of 6.086 km/s of delta-V.
That's actually not a whole lot, in astrodynamics terms. It takes more than that to reach low Earth orbit.
But Ceres is big.
It has a mass of 9.393×10$^{20}$ kg, so changing its velocity by 6.086 km/s would require an impulse of 5.76×10$^{24}$ newton-seconds.
In order for a Hohmann transfer to work, the rocket burns at the beginning and end of the maneuver should ideally be instantaneous. This, of course, is impossible without destroying the asteroid and killing everyone on it; but nuclear pulse propulsion is probably about the closest you'll get without going well beyond the near future.
The engineers of Project Orion concluded that a nuclear pulse drive based on their design could potentially reach a specific impulse of up to 100,000 seconds. Specific impulse, by the way, is a measure of the efficiency of a rocket engine. A specific impulse of 100,000 seconds means that a sufficiently-refined Orion drive could support the weight of its own fuel in Earth gravity for about 100,000 seconds.
The amount of fuel actually required to pull off this maneuver can be derived from the infamous Rocket Equation:
$$\Delta v = I_{sp} \cdot g \cdot \ln \left(\frac {m + m_p} m\right)$$
where $I_{sp}$ is the specific impulse, $g$ is Earth's gravity, $m$ is the mass of Ceres, in this case, and $m_p$ is the mass of the thermonuclear bombs serving as propellant.
Solving this for $m_p$ gives
$$m_p = m \left(e^{\frac {\Delta v} {I_{sp} \cdot g}} - 1\right)$$
Invoking Wolfram|Alpha once again indicates that you'll need 2.72×10$^{18}$ kg of nuclear weapons to start the transfer orbit, and 3.16×10$^{18}$ kg of them at the end. And that's if you get someone to restock your asteroid-ship with the second batch of nukes midway through.
Plus whatever you'd need to actually get the asteroid into orbit around Mars, which depends on how close you want it to orbit.
Good luck!