A body of mass m is taken from the surface of earth to a height...

Initial potential energy of the body at the surface of the earth is given by:

$U_i = -\dfrac{GMm}{R_e}$

Final potential energy of the body at a height $h = 2R_e$ from the surface of the earth is:

$U_f = -\dfrac{GMm}{R_e + h} = -\dfrac{GMm}{R_e + 2R_e} = -\dfrac{GMm}{3R_e}$

The increase in potential energy is:

$\Delta U = U_f - U_i = -\dfrac{GMm}{3R_e} - \left(-\dfrac{GMm}{R_e}\right)$

$\Delta U = \dfrac{GMm}{R_e} \left(1 - \dfrac{1}{3}\right) = \dfrac{2GMm}{3R_e}$

Acceleration due to gravity at the surface of the earth is $g = \dfrac{GM}{R_e^2}$, which gives $GM = gR_e^2$.

Substituting this value, we get:

$\Delta U = \dfrac{2(gR_e^2)m}{3R_e} = \dfrac{2}{3}mgR_e$

Answer: $\dfrac{2}{3}mgR_e$