"For a 380K moon the extravagantly generous upper limit is now 25mW/m2."

Well, Sun apparently is 5780 K instead of 6000 K, so if cubing things that would also make a difference.

"As for the geometry issue, I think you need to study some calculus. It's not as hard as you're trying to make it."

From wiki:
"Temperature of the Earth

Similarly we can calculate the effective temperature of the Earth TE by equating the energy received from the Sun and the energy transmitted by the Earth, under the black-body approximation:

where TS is the temperature of the Sun, rS the radius of the Sun, and a0 is the distance between the Earth and the Sun. Thus resulting in an effective temperature of 6°C on the surface of the Earth."

If you plug in Moon radius and 380 K to the above, one gets around 18 K as compared to the Sun giving 279 K

"Have fun with it, I'm out."

Ok, thanks.
But I do think that something around a watt per sq meter would be a significant amount. One could look at simply from point of view the total amount energy- the total sq meters of earth facing the Moon- a huge number times 1 or say .025.
But I think it's more significant in the likelihood that more of this energy would absorb in a portion of the atmosphere which fairly transparent to the Sun's energy and air temperature is cold with little heat loss in terms of convection.