Solution:

For heights much smaller than the radius ($h \ll R$), $g_h \approx g(1 - \frac{2h}{R})$.nFor depth $d$, $g_d = g(1 - \frac{d}{R})$. Equating the two gives $1 - \frac{2h}{R} = 1 - \frac{d}{R}$.nThus $d = 2h$. Since $h = 1 \text{ km}$, $d = 2 \times 1 = 2 \text{ km}$.