The part of your question about the non-visible light gives quite surprising (to me at least) results! :

Compare the power from the sun with the cosmic microwave background ("CMB"), taking both as blackbody radiators:
The temperatures are 2.725 and 5778 K respectively (wikipedia :))
The solid angle of the CMB is obviously $4\pi$. The diameter of the sun as seen from the earth is 32' (= 32/60 degrees) according to wikipedia, which gives a solid angle of $6.8\cdot 10^{-5}$. This would give a ratio in the angles of $5.4\cdot 10^{-6}$.
(To be precise, we need a factor of 1/2 more here, since the sunlight is absorbed only by half of the earth, whereas the CMB is always "shining". So I will use a ratio of $2.7\cdot 10^{-6}$. Ignore the factor 1/2 if this reasoning seems too complicated)

So we get a ratio of the radiation power of $$\frac {P_\mathrm{sun}}{P_\mathrm{CMB}} = \frac {T_\mathrm{sun}^4}{T_\mathrm{CMB}^4} \cdot \frac {\Omega_\mathrm{sun}}{\Omega_\mathrm{CMB}} = 2\cdot 10^{13} * 2.7\cdot 10^{-6} = 54\,\mathrm{million}$$

Comparing with 125 million from SpiderPig (which is definitely the more precise answer), I get the astonishing result, that the CMB "shines" more than twice as powerful than all the other stars combined!

The part of your question about the non-visible light gives quite surprising (to me at least) results! :

Compare the power from the sun with the cosmic microwave background ("CMB"), taking both as blackbody radiators:
The temperatures are 2.725 and 5778 K respectively (wikipedia :))
The solid angle of the CMB is obviously $4\pi$. The diameter of the sun as seen from the earth is 32' (= 32/60 degrees) according to wikipedia, which gives a solid angle of $6.8\cdot 10^{-5}$. This would give a ratio in the angles of $5.4\cdot 10^{-6}$.
(To be precise, we need a factor of 1/2 more here, since the sunlight is absorbed only by half of the earth, whereas the CMB is always "shining". So I will use a ratio of $2.7\cdot 10^{-6}$. Ignore the factor 1/2 if this reasoning seems too complicated)

So we get a ratio of the radiation power of $$\frac {P_\mathrm{sun}}{P_\mathrm{CMB}} = \frac {T_\mathrm{sun}^4}{T_\mathrm{CMB}^4} \cdot \frac {\Omega_\mathrm{sun}}{\Omega_\mathrm{CMB}} = 2\cdot 10^{13} * 2.7\cdot 10^{-6} = 54\,\mathrm{million}$$

Comparing with 125 million from SpiderPig (which is definitely the more precise answer), I get the astonishing result, that the CMB "shines" more than twice as powerful than all the other stars combined!

PS: the reasoning for the ration of CMB to other stars has some difficulty, of course...
The value for the stars is derived from measurements with visible light, the other value is based on the total power of a blackbody. So the ratio is only correct if the stars have the same temperature as the sun (or more precise, if the ratio of visible power to total power is the same). Since the light of stars is less visible than the sun's (funnily, "per definition" :) - it's the evolutionary definition of visible!) they really produce more power than we think by looking (i.e. the 1 to 125 million of the sun). So the ratio might be (maybe much) smaller than two, but it's still astonishing...