The gravitational potential at a point above the surface of earth...

Gravitational potential at a distance $r$ from the centre of the earth is given by, $V = \dfrac{- GM}{r}$.

Therefore, we can write

$- \dfrac{G{M}_{E}}{{R}_{E} + h} = - 5 . 12 \times {10}^{7} \ldots \left(i\right)$

also acceleration due to gravity at that point will be

$\dfrac{G{M}_{E}}{{\left({R}_{E} + h\right)}^{2}} = 6 . 4 \ldots \left(ii\right)$

By (i) and (ii), we get

$\dfrac{- \dfrac{G{M}_{E}}{{R}_{E} + h}}{\dfrac{G{M}_{E}}{{\left({R}_{E} + h\right)}^{2}}} = \dfrac{5 . 12 \times {10}^{7}}{6 . 4}$

$⇒ {R}_{E} + h = 80 \times {10}^{5}m = 8000km ⇒ h = 8000km - 6400km = 1600km$