Explanation

(d) Escape velocity from the earth
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {(v}_{e}) = 11.2\ {kms}^{- 1}$
Let the mass, radius and density of earth be $M,\ R\ and\ \rho$ respectively and for given planet mass, radius and density are $M^{'},\ R^{'}and\ \rho^{'},$ respectively.
$\therefore$ Escape velocity from the earth
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{\frac{2G \times \left( \frac{4}{3}\pi R^{3}\rho \right)}{R}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{\frac{8G\pi R^{2}\rho}{3}\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Similarly, escape velocity from the given planet
$\ \ \ \ \ \ \ \ \ \ \ \ \ {v'}_{e} = \sqrt{\frac{8G\pi{R'}^{2}\rho}{3}\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
Dividing Eq. (i) by Eq. (ii) we get
$\ \ \ \ \ \ \ \ \ \ \ \frac{v_{e}}{{v^{'}}_{e}} = \sqrt{\frac{8G\pi R^{2}\rho}{3}\ } \times \sqrt{\frac{3}{8G\pi{R^{'}}^{2}\rho}\ } = \sqrt{\frac{R^{2}}{{R^{'}}^{2}}}$
$or\ \ \ \ \frac{11.2}{{v^{'}}_{e\ }} = \frac{R}{R^{'}}$
$or\ \ \ \ \frac{11.2}{{v^{'}}_{e}} = \frac{R}{2R}$
$\therefore\ \ \ \ \ \ \ {v'}_{e} = 22.4\ {kms}^{- 1}$