Explanation

(b) The acceleration due to gravity
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}$
At a height $h$ above the earth’s surface, the acceleration due to gravity is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} = \frac{GM}{{(R + h)}^{2}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{g}{g^{'}} = \left( 1 + \frac{h}{R} \right)^{- 2} = \left( 1 + \frac{h}{R} \right)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{g^{'}}{g} = \left( 1 + \frac{h}{R} \right)^{- 2} = \left( 1 - \frac{2h}{R} \right)$
$but\ \ \ \ \ \ \ \ \ \ \ \ g^{'} = \frac{g}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (given)$
$\therefore\ \ \ \ \ \ \ \ \ \ \frac{g/2}{g} = 1 - \frac{2h}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{2h}{R} = \frac{1}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ h = \frac{R}{4}$