Explanation

(c) Speed of electron in $n^{th}$ orbit of hydrogen atom $v = \frac{e^{2}}{2\varepsilon_{0}nh}$
In ground state $n = 1 \Rightarrow v = \frac{e^{2}}{2\varepsilon_{0}h}$
$\Rightarrow \frac{v}{c} = \frac{e^{2}}{2\varepsilon_{0}ch} = \frac{\left( 1.6 \times 10^{- 19} \right)^{2}}{2 \times 8.85 \times 10^{- 12} \times 3 \times 10^{8} \times 6.6 \times 10^{- 34}}$
$= \frac{1}{137}$