Answer:
T=270 K
Step-by-step explanation:
Power absorbed by the Moon
![P_(absorbed)= [1-0.12]solar\ constant*\pi R^2](https://img.qammunity.org/2020/formulas/physics/college/dmwj5zqt3hjq13ypg3fzokveqoqi4w259s.png)
the reason for choosing the area as π R^2 and not 4π R^2 is because the absorbed light is only from the surface facing the Sun which is a disc.
Using Stefan Boltzmann's law,
![P_(emitted)= \sigma T^4(4\pi R^2)](https://img.qammunity.org/2020/formulas/physics/college/2f3vgexrswk50fmqja0iftp1nwq0oraewz.png)
So, using Conservation of Energy,
![[1-0.12]solar\ constant*\pi R^2<strong>=</strong>\sigma T^4(4\pi R^2)](https://img.qammunity.org/2020/formulas/physics/college/ex085c3vr0wrzvmocqufjem87vx5z7f1vq.png)
this gives, T = 270 K. This is the temperature of the sunlight side of the Moon.