The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the…

   Acceleration due to gravity in both cases are

$\begin{array}{c}\mathrm{Above}\mathrm{ }\mathrm{earth}\mathrm{ }\mathrm{surface}\\ g^{\prime }=g\left(1-\frac{2h}{R_e}\right)\end{array}$

$\begin{array}{c}\mathrm{Below}\mathrm{ }\mathrm{earth}\mathrm{ }\mathrm{surface}\\ g^{\prime }=g\left(1-\frac{d}{R_e}\right)\end{array}$
As per the problem, $g\left(1-\frac{2h}{R_e}\right)=g\left(1-\frac{d}{R_e}\right)$

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

The depth is $d=2\times 1\mathrm{km}$