b
\(\begin{array}{l}
Speed\,on\,reaching\,ground\,v = \sqrt {{u^2} + 2gh} \\
Now,\,v = u + at\\
\Rightarrow \,\,\,\,\sqrt {{u^2} + 2gh} = - u + gt\\
Time\,taken\,to\,reach\,highest\,{\rm{point}}\,is\,t = \frac{u}{g},\\
\Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} = \frac{{nu}}{g}\left( {from\,question} \right)\\
\Rightarrow 2gH = n\left( {n - 2} \right){u^2}
\end{array}\)
