(b)
KE$= \frac{3}{2}\ kT$
Where, $k$ is constant.
$KE\ \propto T$
Here the temperature is same. Hence, for 1 g of $H_{2}$ and 1 g of ${CH}_{4}$ which are taken in two vessels, of 1 L each at same temperature, the kinetic energy per mole will be the same.