(a) According to kinetic interpretation of temperature
$K.E. = \left( \frac{1}{2}mv^{2} \right) = \frac{3}{2}kT$
$\Rightarrow 10.2 \times 1.6 \times 10^{- 19} = \frac{3}{2} \times \left( 1.38 \times 10^{- 23} \right)T$
$\Rightarrow T = 7.9 \times 10^{4}K$