What happens to the internal energy of an object as it radiates more energy than it absorbs?

If you just mean the blackbody radiation, then that the radiated power both depends on the object's surface area, $A$, and its temperature to the forth power, $T^4$, according to $P_{\rm rad} = \epsilon\sigma A T^4$, where $\epsilon$ is the object's emissivity (material dependent) and $\sigma=5.67\times10^{-8}{\rm\,W\,m^{-2}\,K^{-4}}$ is the Stefan-Boltzmann constant. Hence it's the surface area which determines the radiation.

Since two objects can have the same volume but vastly different surface areas (think of a sheet of paper which has a large surface area but a very small volume), the thermal radiation is not dependent on the volume. However if you have two objects of the same shape, e.g. two cubes or two spheres, then the surface area scales with the volume as $A\propto V^{2/3}$, which would then also affect the thermal radiation, but only for two object of the same shape.

To answer your final question: If you have two objects at the same temperature and material, then the one with the larger surface area will radiate more energy.