The gravitational potential energy ($U$) of a body of mass $m$ at a distance $r$ from the centre of the Earth is given by $U = -\frac{GMm}{r}$.

1. Initial State: At the Earth's surface, $r_1 = R$. So, $U_i = -\frac{GMm}{R}$.
2. Final State: At a height $h = 2R$ above the surface, the distance from the centre is $r_2 = R + h = R + 2R = 3R$. So, $U_f = -\frac{GMm}{3R}$.
3. Change in Potential Energy ($\Delta U$): $\Delta U = U_f - U_i = \left( -\frac{GMm}{3R} \right) - \left( -\frac{GMm}{R} \right)$ $\Delta U = \frac{GMm}{R} \left( 1 - \frac{1}{3} \right) = \frac{2GMm}{3R}$
4. Substitution: We know that acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$, which implies $GM = gR^2$. Substituting this into the equation: $\Delta U = \frac{2(gR^2)m}{3R} = \frac{2}{3}mgR$.