Explanation

(a) Let $R$ be the radius of earth and $\rho$ its density, then since shape of earth is assumed spherical
we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass\ of\ earth = volume \times density$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = \frac{4}{3}\pi R^{3} \times \rho\ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
The acceleration due to gravity which arises in the body due to gravitational force of attraction is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
Putting the value of $M$ from Eq.(i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ g = \frac{G\frac{4}{3}\pi R^{3}\rho}{R^{2}} = G\frac{4}{3}\pi R\rho\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(iii)$
$Given,\ \ \ \ \ \rho_{p} = \rho,\ R_{p} = 0.2R_{e}$
$\therefore\ g_{p} = G\frac{4}{3}\pi R_{p}\rho_{p} = G \times \frac{4}{3}\pi \times 0.2R\rho = 0.2g$