A uniformly charged hemisphere of radius b and charge density $\rho$  has a hemispherical cavity of radius $a(a=\frac{b}{2})$ cut from its centre.  If the potential at the centre of the cavity is  $\frac{n\rho b^2}{16{\epsilon }_0}$, then n= ?

Potential at the centre of a hemispherical shell  $=\frac{kQ}{R}$
Lets divide the hemisphere into thin hemispherical shells. Lets take a hemispherical shell of radius r and thicnkness  $dr$.
Potential at its centre $=\frac{kdq}{r}$ 
 $$\begin{gathered} dV=\frac{k\left({\rho }^2{\pi }^{2}dr\right)}{r} \\ \int dV=\frac{\rho }{2{\epsilon }_0}{\int }_a^{b}rdr \\ V=\frac{\rho }{2{\epsilon }_0}{\left[\frac{r^2}{2}\right]}_a^b =\frac{\rho }{4{\epsilon }_0}(b^2-\frac{b^2}{4})=\frac{3\rho b^2}{16{\epsilon }_0} \end{gathered}$$