At a certain depth d below surface of earth, value of acceleration due to gravity becomes four times that of…

The magnitude of acceleration due to gravity $g_1$ at a depth of $d$ below the surface of Earth is

$g_1=G\frac{M}{R^3}\left(R-d\right)=G\frac{M}{R^2}\left(1-\frac{d}{R}\right)$    ....(1)

The magnitude of acceleration due to gravity $g_2$ at a height $d$ above the surface of Earth is

$g_2=G\frac{M}{{\left(R+d\right)}^2}$    ....(2)

Given here, $g_1=4g_2$

$\frac{GM}{R^2}\left(1-\frac{d}{R}\right)=\frac{4GM}{{\left(R+3R\right)}^2}$

$\Rightarrow \frac{GM}{R^2}\left[1-\frac{d}{R}\right]=\frac{4GM}{{\displaystyle {\left(4R\right)}^2}}$

$$\begin{gathered} \Rightarrow 1-\frac{d}{R}=\frac{1}{4} \\ \Rightarrow \frac{d}{R}=\frac{3}{4} \\ \Rightarrow d=\frac{3}{4}R \end{gathered}$$

$\Rightarrow d=4800 \mathrm{km}$