Explanation

(c) The average kinetic energy of monoatomic gas molecule is $K = \frac{3}{2}k_{B}T$
Where $k_{B}$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvin
$K = \frac{3}{2} \times \left( 1.38 \times 10^{- 23}JK^{- 1} \right) \times (300K)$
$= \frac{3 \times (1.38 \times 10^{- 23}JK^{- 1}) \times (300K)}{2 \times (1.6 \times 10^{- 19}J/eV)}$
$= 3.9 \times 10^{- 2}eV = 0.039eV$