At a certain depth d below surface of earth, value of acceleration...

The magnitude of acceleration due to gravity ${g}_{1}$ at a depth of $d$ below the surface of Earth is

${g}_{1} = G\dfrac{M}{{R}^{3}}\left(R - d\right) = G\dfrac{M}{{R}^{2}}\left(1 - \dfrac{d}{R}\right)$ ....(1)

The magnitude of acceleration due to gravity ${g}_{2}$ at a height $d$ above the surface of Earth is

${g}_{2} = G\dfrac{M}{{\left(R + d\right)}^{2}}$ ....(2)

Given here, ${g}_{1} = 4{g}_{2}$

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

$⇒ \dfrac{GM}{{R}^{2}}\left[1 - \dfrac{d}{R}\right] = \dfrac{4GM}{{\left(4R\right)}^{2}}$

$⇒ 1 - \dfrac{d}{R} = \dfrac{1}{4} ⇒ \dfrac{d}{R} = \dfrac{3}{4} ⇒ d = \dfrac{3}{4}R$

$⇒ d = 4800km$