Explanation

(a) Radius of earth $R = 6400\ km\ \therefore h = \frac{R}{4}$
Acceleration due to gravity at a height $h$
$g_{h} = g\left( \frac{R}{R + h} \right)^{2} = g\left( \frac{R}{R + \frac{R}{4}} \right)^{2} = \frac{16}{25}g$
At depth $'d'$ value of acceleration due to gravity
$g_{d} = \frac{1}{2}g_{h}$ (According to problem)
$\Rightarrow g_{d} = \frac{1}{2}\left( \frac{16}{25} \right)g \Rightarrow g\left( 1 - \frac{d}{R} \right) = \frac{1}{2}\left( \frac{16}{25} \right)g$
By solving we get $d = 4.3 \times 10^{6}m$