At what depth below surface of the earth, the acceleration due to gravity will be half of its value that at 1600km above the surface of the earth ?

Given, Height, H=1600km
Let depth be d
Acceleration due to gravity at height, depth and surface be  $g_h,g_{d}andg.$
Also,  $g_d=\frac{g_h}{2}\mathrm{..}\left(i\right)$
As we know that, $g_h=g(1-\frac{2h}{R})$
 $\Rightarrow g_h=g(1-\frac{2\times 1600}{6400})=\frac{g}{2}and{g}_d=g(1-\frac{d}{R})$
From Eq. (i)
$\frac{1}{2}\times \frac{g}{2}=g(1-\frac{d}{R})\Rightarrow \frac{d}{R}=\frac{3}{4}\Rightarrow d=\frac{3}{4}R=\frac{3}{4}\times 6400=4.8\times {10}^{6}m$