The radius of the earth is R and acceleration due to gravity at its surface is g. If a body of mass m is sent to a height of $\frac{\text{R}}{\text{4}}$ from the earth’s surface, the potential energy is

$\text{P}\text{.E}=\frac{\text{Mgh}}{\text{1}+\frac{\text{h}}{\text{R}}}$                        $\text{h}=\frac{\text{R}}{\text{4}}$

Note : here P.E is referring the gravitational potential energy possessed by the body from surface of earth not from centre of earth

P.E & G.P.E are not same.      

$\text{P}\text{.E}=\frac{\text{Mg}\left(\frac{\text{R}}{\text{4}}\right)}{\text{1}+\frac{\frac{\text{R}}{\text{4}}}{\text{R}}}$  

$\frac{\text{MgR}}{\text{4}\left(\text{1+}\frac{\text{<menclose notation="updiagonalstrike">R </menclose>}}{\text{4<menclose notation="updiagonalstrike">R </menclose>}}\right)}$  

 $=\frac{\text{mgR}}{4\left(1+\frac{1}{4}\right)}=\frac{\text{mgR}}{\cancel{4}\left(\frac{5}{\cancel{4}}\right)}=\frac{\text{mgR}}{5}$