(b) Gravitational pull depends upon the acceleration due to gravity on that planet
$M_{m} = \frac{1}{81}M_{e},\ g_{m} = \frac{1}{6}g_{e}$
$g = \frac{GM}{R^{2}} \Rightarrow \frac{R_{e}}{R_{m}} = \left( \frac{M_{e}}{M_{m}} \times \frac{g_{m}}{g_{e}} \right)^{1/2} = \left( 81 \times \frac{1}{6} \right)^{1/2}$
$\therefore R_{e} = \frac{9}{\sqrt{6}}R_{m}$