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Variation of g with height and depth
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The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then
For small h, equal g requires d = 2h.
Above the earth's surface: $g' = g\left(1 - \frac{2h}{R_e}\right)$, so $\Delta g = g\frac{2h}{R_e}$ ...(1) Below the earth's surface: $g' = g\left(1 - \frac{d}{R_e}\right)$, so $\Delta g = g\frac{d}{R_e}$ ...(2) From (1) and (2): $d = 2h = 2\times1$ km = 2 km