The acceleration due to gravity at a height 1 km above the earth is...

Acceleration due to gravity in both cases are

$$\begin{array}{ll}Aboveearthsurface \\ {g}^{′} = g\left(1 - \dfrac{2h}{{R}_{e}}\right)\end{array}$$

$$\begin{array}{ll}Belowearthsurface \\ {g}^{′} = g\left(1 - \dfrac{d}{{R}_{e}}\right)\end{array}$$
As per the problem, $g( 1 - \dfrac{2h}{{R}_{e}} ) = g( 1 - \dfrac{d}{{R}_{e}} )$

Simplifying the above relations, we get
$d = 2h$

The depth is $d = 2 \times 1 km$