At what distance above and below the surface of the earth a body...

The acceleration due to gravity at a height $h$ above the surface of the Earth is given by

${g}_{p} = \dfrac{g{R}^{2}}{( R + h{)}^{2}} . . . \left(1\right)$

The acceleration due to gravity at a depth $h$ below the surface of the Earth is given by

${g}_{q} = g\left(1 - \dfrac{h}{R}\right) . . . \left(2\right)$

According to the given problem,

${g}_{p} = {g}_{q} . . . \left(3\right)$

Equations (1), (2) and (3) imply that

$\dfrac{g}{{\left(1 + \dfrac{h}{R}\right)}^{2}} = g\left(1 - \dfrac{h}{R}\right) ⇒ \left(1 - \dfrac{{h}^{2}}{{R}^{2}}\right)\left(1 + \dfrac{h}{R}\right) = 1 . . . \left(4\right)$

Take $\dfrac{h}{R} = x$

So, from equation (4), it follows that

${x}^{3} - x + {x}^{2} = 0 ⇒ {x}^{2} + x - 1 = 0 ⇒ x = \dfrac{\sqrt{5} - 1}{2} ⇒ h = \dfrac{R}{2} ( \sqrt{5} - 1 )$