(d) Average kinetic energy
$E = \frac{3}{2}kT \Rightarrow \frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} = \frac{(273 - 23)}{(273 + 227)} = \frac{250}{500} = \frac{1}{2}$
$\Rightarrow E_{2} = 2E_{1} = 2 \times 5 \times 10^{- 14} = 10 \times 10^{- 14}erg$