The ratio of the weights of a body on Earth’s surface to that...

Weight $= m\left(\dfrac{GM}{{R}^{2}}\right)$
$∴ \dfrac{9}{4} = \dfrac{G{M}_{e}}{{R}_{e}^{2}} \times \dfrac{{R}_{p}^{2}}{G{M}_{p}}$
Given that, ${M}_{p} = \dfrac{{M}_{e}}{9}$
$∴ \dfrac{9}{4} = \dfrac{{M}_{e}}{\left(\dfrac{{M}_{e}}{9}\right)} \times \dfrac{{R}_{p}^{2}}{{R}_{e}^{2}}$
${R}_{p}^{2} = {R}_{e}^{2} \times \dfrac{1}{4}$
${R}_{p} = \dfrac{{R}_{e}}{2}$
As, ${R}_{e} = R$
$⇒ {R}_{p} = \dfrac{R}{2}$