Explanation

(d) Let escape velocity be$\ v_{e}$, then kinetic energy is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}{mv}_{e}^{2}\ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$and\ escape\ energy = + \frac{{GM}_{e}m}{R_{e}}\ \ \ \ \ \ldots(ii)$
Equating Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}m_{e}^{2} = \frac{{GM}_{e}m}{R_{e}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{\frac{2{GM}_{e}}{R_{e}}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = \frac{{2GM}_{e}}{v_{e}^{2}}$
Given, $\ \ \ \ \ \ \ G = 6.67 \times 10^{- 11}N–m^{2}/kg,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_{e} = 6 \times 10^{24}kg,\ v_{e} = 3 \times 10^{8}\ {m/s}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{(3 \times 10^{8})^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 8.89 \times 10^{- 3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R \approx 9 \times 10^{- 3}m = 9mm$