Explanation

(c) Acceleration due to gravity at a height $h$ above the earth’s surface is $g_{h} = \frac{g}{\left( 1 + \frac{h}{R} \right)^{2}}$
Where $g$ is the acceleration due to gravity on the earth’s surface
At $h = \frac{R}{2},\ g_{h} = \frac{g}{\left( 1 + \frac{R}{2R} \right)^{2}} = \frac{4g}{9}$
At $h = R,\ g_{h} = \frac{g}{\left( 1 + \frac{R}{R} \right)^{2}} = \frac{g}{4}$
Acceleration due to gravity at a depth $d$ below the earth’s surface is $g_{d} = g\left( 1 - \frac{d}{R} \right)$
At $d = \frac{R}{2},\ g_{d} = g\left( 1 - \frac{2}{2R} \right) = \frac{g}{2}$
At the centre of earth, $d = R$
$g_{d} = g\left( 1 - \frac{R}{R} \right) = 0$
Thus, the acceleration due to gravity is maximum on the earth’s surface