At what depth below the surface of the ear

Radius of earth $R = 6400\, km $

$h = \frac{R}{4}$ 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) 

$⇒$ ${g_d} = \frac{1}{2}\left( {\frac{{16}}{{25}}} \right)g$ $ \Rightarrow g\left( {1 - \frac{d}{R}} \right)$

$ = \frac{1}{2}\left( {\frac{{16}}{{25}}} \right)\;g$

By solving we get $d = 4.3 \times {10^6}\,m$