Work done = increase in gravitational potential energy
$W_1 =\large\frac{mg R}{1+\Large\frac{R}{R}}=\large\frac{mgR}{2}$
$\bigg[ \Delta u =\large\frac{mgh}{1+\Large\frac{h}{R}}\bigg]$ Where 'h' height above the earth surface
Given $2W_2=W_1$
Where $W_2$ = work done in lifting
through a height 'h'
$\large\frac{mgR}{2} =\frac{2mgh}{1+\Large\frac{h}{R}}$
Solving we get,
$h= \large\frac{R}{3}$
Hence b is the correct answer.