Decrease in Kinetic energy = increase in gravitational potential energy
$\large\frac{1}{2}$$ mv^2= \Delta v$
$\large\frac{1}{2}$$ mv^2=\large\frac{mgh}{1+\Large\frac{h}{R}}$ where h is the height above earth
$\large\frac{1}{2}$$ mv^2=\large\frac{mgnR}{1+\Large\frac{nR}{R}}$
Solving we get,
$v^2= \large\frac{2gnR}{n+1}$
$v= \sqrt {\large\frac{2 ngR}{n+1}}$
Hence a is the correct answer.