A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R ,…

According to the Gauss law, the total flux linked with a closed surface is $\frac{1}{{\epsilon }_0}$ times the charge enclosed by the closed surface. 

As we can see from the diagram, the given shape encloses the charge $q$, therefore from Gauss law, we can write total flux $=\frac{q}{{\epsilon }_0}$.

Now, the plane connecting hemispherical shape & conical shape cuts the charge in two equal parts, therefore flux through the conical part will be exactly half of the total flux value.

Hence, flux through conical part is, $\frac{q}{2{\epsilon }_0}$.

Therefore,

$n=3$