Where does that 3-10 m^2 per kW estimate come from? The best I could guess based on my shaky grasp of radiation is to equate internally generated power to total radiated power using the Stefan-Boltzmann Law:
Based on that, you need 3-10 m^2 of radiative surface area per kilowatt, but that's assuming your equilibrium temperature is 203 to 275 K (-70 to 2 °C). Assuming I haven't made some basic mistake, couldn't you decide to heat part of your surface to some much higher temperature and radiate most of your internal power out of that part?
This is what our current spacecraft radiators manage. You also have to take into account that radiators also interact with the sun and with the rest of the spacecraft.
Of course, you could have a heat pump pushing heat to a higher temperature surface, some current radiators do that, but it has its limitations requires more energy for the pump the higher temperature difference you want to sustain.
https://www.wolframalpha.com/input/?i=1+kW%2F%28Stefan-Boltz...
Based on that, you need 3-10 m^2 of radiative surface area per kilowatt, but that's assuming your equilibrium temperature is 203 to 275 K (-70 to 2 °C). Assuming I haven't made some basic mistake, couldn't you decide to heat part of your surface to some much higher temperature and radiate most of your internal power out of that part?