Acceleration due to gravity at the surface of earth
$ {{g}_{h}}=g{{\left( 1+\frac{h}{R} \right)}^{-2}} $
$ {{g}_{h}}=g\left( g-\frac{2h}{R} \right) $
Acceleration due to gravity at depth d from the surface of the earth
$ {{g}_{d}}=g\left( 1-\frac{d}{R} \right) $
Given: $ {{g}_{h}}={{g}_{d}} $
$ \therefore $ $ \frac{2h}{R}=\frac{d}{R} $
Or $ d=2h $ Or $ d=10\,km $