Explanation

(b) The acceleration due to gravity on the new planet can be using the relation
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$but\ M = \frac{4}{3}\pi R^{3}\rho,\rho\ being\ density.$
Thus, Eq. (i) becomes
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{G \times \frac{4}{3}\pi R^{3}\rho}{R^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = G \times \frac{4}{3}\pi R\rho$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ g \propto R$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \frac{g^{'}}{g} = \frac{R^{'}}{R}$
$\Longrightarrow \ \ \ \ \ \ \ \ \frac{g^{'}}{g^{'}} = \frac{3R}{R} = 3$
$\Longrightarrow \ \ \ \ \ \ \ \ g^{'} = 3g$