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

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

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 $\mathrm{d}=4.3\times {10}^6\mathrm{m}$