As a rule of thumb, you get about 1 kW of sunlight per meter squared on the earths surface. In space you have about 1.3 kW per meter squared without the atmosphere. So ignoring things like clouds, the precise latitude, time of year, time of day etc. You can say about 30% of the energy doesn't reach the earth.
The temperature rock melts varies a lot, but silicas, and some of the more heat resistant minerals melt in the 1200-1300 C range.
The size of of the Odeillo tower is not as important as knowing the total area of the heliostats that they are using to direct light to the parabolic mirror and the area of the target that is illuminated. In that case they have 63 heliostats that track the sun and total area of about 2835 square meters. Roughly you can say that it is magnifying the sun about 3000 times, and they rate the facility at about 1 megawatt. It seems it can reach temperatures of about 3,000-3500 C, although the details are not that clear.
By the way, there a some theoretical limits as to how much you can concentrate sunlight, and the record is somewhere around 84,000 times. However, that technique isn't very helpful for this application. More generally for a terrestrial solar furnace a magnification of 2000, is pretty reasonable.
Well, you have 1 km diameter space ships that you want to position. The best you can do in directing the light as a beam, (assuming the light from the sun is mostly parallel before it hits the space ship, would be diffraction limited by the angle theta=1.22*wavelength_in _meters/1000meters. This angle is pretty small and is in radians, 6.1xe-10, but if in geosynchronous orbit about 35,760 km then the 1km spot would be 70 meters larger, so that that sounds still about a km. But if as far out at the moon, 363,300 km, then the spot would have a diameter would be 1.72 km in diameter. This would depend on the wavelength with shorter wavelengths spread out more. To make your spots smaller you could put a slight curvature on your 1 km spaceships, but you will still end up being limited by diffraction by the diameter of your space ship.
So to get the energy density concentrating 2000 times, you could use 2000 spaceships and overlap the spots. Then to get the line to scan, you could add them in increments of 2000, each increment adding km to the line. Or less if you want to overlap some for efficiency.
The circumference of the earth is about 40,000 km, so 1/2 that is 20,000 so 20,000 spaceships in lined up in groups of 2000 would be one option. You might think that you could just let the earth rotate underneath the space ship array. Well, then the equator is moving faster than the higher latitudes 40,000/24 hrs is about 1,667 km/hr, so that is a minor problem.... you would be under the spot for about 18 seconds, and might not be quite hot enough to melt the rock at the equator, as you go up in latitude you would spend more time within the spot.
Of course everything up to now neglected the atmosphere, or the oceans, or vegetation. This gets sort of complicated. Initially, assuming turning on one beam from the 2000 spaceships, you will have absorption from the atmosphere, with the atmosphere warming up in the spot, that will create convection. You will also get some changes in the refractive index and a variety of scintillation effects, so the way the light actually reaches the ground will be distorted and changing due to convection and the turbulence. The molten rock of course will also create convection currents and storms. The burning material will be carried up into the atmosphere and change the reflectivity the way the light is absorbed in the atmosphere.
However, water vapor and cloud formation will likely cause the biggest effects. How this might work could be complicated. Usually, one would think that the clouds would be very reflective so much less light would reach the ground. However, there may be enough energy to vaporize water droplets, perhaps punching a hole through the clouds. Then there are secondary effects like greenhouse gas effects...
