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

Radius of earth R = 6400 km $\therefore \mathrm{h}=\frac{\mathrm{R}}{4}$

Acceleration due to gravity at a height h

${\mathrm{g}}_{\mathrm{h}}=\mathrm{g}{\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2=\mathrm{g}{\left(\frac{\mathrm{R}}{\mathrm{R}+\frac{\mathrm{R}}{4}}\right)}^2=\frac{16}{25}\mathrm{g}$

At depth 'd' value of acceleration due to gravity

 ${\mathrm{g}}_{\mathrm{d}}=\frac{1}{2}{\mathrm{g}}_{\mathrm{h}}$       (According to problem)

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

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