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

Solution :
$$\eqalign{ & {g_h} = {\text{Acceleration due to gravity at height }}h{\text{ above earth's surface}} \cr & = g{\left( {\frac{R}{{R + h}}} \right)^2} = g\left( {1 - \frac{{2h}}{R}} \right) \cr & {g_d} = {\text{Acceleration at depth }}d{\text{ below earths surface}} \cr & = g\left( {1 - \frac{d}{R}} \right) \cr & {\text{Given, when }}h = 1\,km,\,{g_d} = {g_h} \cr & {\text{or}}\,\,g\left( {1 - \frac{d}{R}} \right) = g\left( {1 - \frac{{2h}}{R}} \right) \cr & \Rightarrow d = 2h\,{\text{or}}\,\,d = 2\,km \cr} $$