A body of mass $m$ is taken from the earth's surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of the body will be:

The gravitational potential energy at a distance $r$ from the center of the earth is $U=-\frac{GMm}{r}$.At the surface: $r_1=R$ $
$U_1=-\frac{GMm}{R}$$Atheight$ h = 2R$:$ r_2 = R + 2R = 3R
$U_2 = -\frac{GMm}{3R}
Change in potential energy $\Delta U=U_2-U_1$:
$$\Delta U=-\frac{GMm}{3R}-\left(-\frac{GMm}{R}\right)=\frac{GMm}{R}-\frac{GMm}{3R}=\frac{2GMm}{3R}$$
Since $g=\frac{GM}{R^2}⟹GM=g{R}^2$:
$$\Delta U=\frac{2(g{R}^2)m}{3R}=\frac{2}{3}mgR$$
Alternatively, using the formula $\Delta U=\frac{mgh}{1+h/R}$:
$$\Delta U=\frac{mg(2R)}{1+2R/R}=\frac{2mgR}{1+2}=\frac{2}{3}mgR$$