It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.
Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler.
You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency.
Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.
And again: by that same law, it just passes right back out again because the same NET amount of radiative power that crosses the boundary and intercepts the smaller sphere is either reflected, transmitted, or scattered. (Since we are discussing diffuse gray bodies here, we can consider it all reflected or scattered because there is no transmissivity.) The radiation that crosses the boundary that does not strike the smaller sphere due to view factor also just passes right back out. You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up.
Once again, no. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
Just NO. Net heat transfer is ALL from hotter to colder, by (e*s) * (Ta^4 - Tb^4).
Let me put it another way: we can easily show how you have gotten your thermodynamics backward by referring to a question you asked earlier. You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F.
The answer is YES, and here is why:
You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does.
The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area).
This clearly illustrates your ass-backward thermodynamic thinking. The radiative power output of the heat source does not change due to the temperature of the walls. At all. The only thing that changes as the wall temperature changes is the heat transfer, which would lessen as you brought up the temperature of the walls. But that isn't because the heat source is using less power, it is because you are putting more power into raising the wall temperature. You are creating a more thermodynamically energetic environment, and that requires power.
Just like your other arguments: you invent power in out of thin air, and claim you can do that because it's "moving" in the opposite direction in which heat transfer is actually taking place.
You are giving physicists a bad name, and I repeat that I am going to show this to all the world to see.