The potential energy of a particle at a distance $r (>R_e)$  from the centre of Earth is,

$U=-G\frac{M_{e}m}{r}$

Initially, the particle is at a height $R_e$ from the surface. Therefore, $r=R_e$.

$\therefore U_i=-G\frac{M_{e}m}{R_e}$

When height is increased to $2R_e$ from surface, $r=2R_e+R_e=3R_e$

$\therefore U_f=-G\frac{M_{e}m}{3R_e}$

Thus, the change in potential energy is

$$\begin{gathered} ∆U=U_f-U_i \\ =-G\frac{M_{e}m}{3R_e}+G\frac{M_{e}m}{R_e} \\ =\frac{2}{3}G\frac{M_{e}m}{R_e} \\ =\frac{2}{3}G\frac{M_e}{{R_e}^2}m{R}_e \\ =\frac{2}{3}mg{R}_e \left(\text{as} g=G\frac{M_e}{{R_e}^2}\right) \end{gathered}$$