A body of mass is taken from earth surface to the height h equal to...

The potential energy of a particle at a distance $r ( > {R}_{e} )$ from the centre of Earth is,

$U = - G\dfrac{{M}_{e}m}{r}$

Initially, the particle is at a height ${R}_{e}$ from the surface. Therefore, $r = {R}_{e}$.

$∴ {U}_{i} = - G\dfrac{{M}_{e}m}{{R}_{e}}$

When height is increased to $2{R}_{e}$ from surface, $r = 2{R}_{e} + {R}_{e} = 3{R}_{e}$

$∴ {U}_{f} = - G\dfrac{{M}_{e}m}{3{R}_{e}}$

Thus, the change in potential energy is

$∆ U = {U}_{f} - {U}_{i} = - G\dfrac{{M}_{e}m}{3{R}_{e}} + G\dfrac{{M}_{e}m}{{R}_{e}} = \dfrac{2}{3}G\dfrac{{M}_{e}m}{{R}_{e}} = \dfrac{2}{3}G\dfrac{{M}_{e}}{{{R}_{e}}^{2}}m{R}_{e} = \dfrac{2}{3}mg{R}_{e}\left(\text{as}g = G\dfrac{{M}_{e}}{{{R}_{e}}^{2}}\right)$