A body of mass $$m$$ is taken from the surface of the earth to a height equal to twice the radius of earth$$(R_e)$$. The increase in potential energy will be  :
(g is acceleration due to gravity at the surface of earth) 

The initial potential energy of the body at the surface of the earth is:

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

The body is taken to a height $$h=2R_e$$.

The final distance of the body from the center of the earth is:

$$r=R_e+h=R_e+2R_e=3R_e$$

The final potential energy of the body at this height is:

$$U_f=-\frac{GMm}{3R_e}$$

The increase in potential energy ($$\Delta U$$) is the difference between the final and initial potential energies:

$$\Delta U=U_f-U_i$$

$$\Delta U=\left(-\frac{GMm}{3R_e}\right)-\left(-\frac{GMm}{R_e}\right)$$

$$\Delta U=\frac{GMm}{R_e}-\frac{GMm}{3R_e}$$

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

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

$$\Delta U=\frac{2GMm}{3R_e}$$

$$g=\frac{GM}{R_e^2}\implies GM=gR_e^2$$

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

$$\Delta U=\frac{2}{3}mgR_e$$