a
(a) 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 \(d = 4.3 \times {10^6}\,m\)