An object of mass 1kg is taken to a height from the surface of...

Gravitational potential energy on the earth surface ${U}_{i} = \dfrac{- GMm}{R}$

Gravitational potential energy at a height $h$ above the earth's surface ${U}_{f} = - \dfrac{GMm}{R + h} = - \dfrac{GMm}{R + 3R} = - \dfrac{GMm}{4R}$

Gain in gravitational potential energy

$\DeltaU = {U}_{f} - {U}_{i} = - \dfrac{GMm}{4R} + \dfrac{GMm}{R} = \dfrac{GMm + 4GMm}{4R} = \dfrac{3GMm}{4R}$

Gravitational potential energy in terms of acceleration due to gravity is $∆ U = \dfrac{3}{4}mgR$

$= \dfrac{3}{4} \times 1 \times 10 \times 64 \times {10}^{5}$

$= 48MJ$