(c) For earth, $g = \frac{GM}{R^{2}} = \frac{4}{3}\pi R\rho G$
For the planet, $g_{1} = \frac{GM_{1}}{R_{1}^{2}} = \frac{4}{3}\ \pi R_{1}\rho G$
$\frac{g}{g_{1}} = \frac{R}{R_{1}} = \frac{6400}{320} = 20$
Let $h$ and $h_{1}$ be the distance upto which the man can jump on surface of the earth and planet, then
$mgh = mg_{1}h_{1}$
$\therefore\ h_{1} = \frac{g}{g_{1}}\ h = 20 \times 5 = 100\ m$