d
average kinetic energy
=$E = \frac{3}{2}kT$
$\Rightarrow$ $\frac{{{E_1}}}{{{E_2}}} = \frac{{{T_1}}}{{{T_2}}} = \frac{{(273 - 23)}}{{(273 + 227)}} = \frac{{250}}{{500}} = \frac{1}{2}$
$\Rightarrow$ ${E_2} = 2{E_1} = 2 \times 5 \times {10^{ - 14}} = 10 \times {10^{ - 14}}\,erg$