c
Given,
$E _{ O _{2}}=0.048 \; eV$
We know that,
Translational kinetic energy,
$E =\frac{3}{2} KT$
where, $k =$ Boltzmann constant
$T =$ Temperature
$\therefore E \propto T$
Since, the temperature is same for both oxygen and nitrogen therefore, $E _{ O _{2}}= E _{ N _{2}}$
$\therefore E _{ N _{2}}=0.048 \; eV$