At what depth below the surface of earth, value of acceleration due to gravity is same as the value at height $h=R$, where $R$ is the radius of earth.

Given that,$g^{\text{'}}$ at depth $d=g^{\text{'}}$ at height $h=R$

$\therefore g\left(1-\frac{d}{R}\right)=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

Substituting, $h=R$, we get

$d=\frac{3}{4}R$

From the graph we can see that value of $g^{\text{'}}$at depth $d=\frac{3}{4}R$ and height $h=R$ are same.