Depth from the surface of the earth at which is acceleration due to gravity is 25% of acceleration due to gravity at the surface

The correct option is D 4800 km
g at the surface is given as $g_s=\frac{4}{3}\pi GR\rho$
and at a depth d it is given as $g_d=\frac{4}{3}\pi G(R-d)\rho$
Thus $\frac{g_d}{g_s}=\frac{R-d}{R}$
Thus for the depth where acceleration is 25% of the surface gravity we get $g_d$ as $\frac{g}{4}$
or
$\frac{g}{4}=g(1-\frac{d}{R})\Rightarrow d=\frac{3}{4}\times R=4800km$ (R is 6371 km)