From Gauss law,

$\phi_{\text {hemisplere }}+\phi_{\text {Cone }}=\frac{ q }{\varepsilon_0}$

Total flux produced from $q$ in $\alpha$ angle

$\phi=\frac{ q }{2 \varepsilon_0}[1-\cos \alpha]$

For hemisphere, $\alpha=\frac{\pi}{2}$

$\phi_{\text {hemisphere }}=\frac{ q }{2 \varepsilon_0}$

From equation $(i)$

$=\frac{ q }{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{ q }{\varepsilon_0}$

$\phi_{\text {cone }}=\frac{ q }{2 \varepsilon_0}$

$\frac{4 q }{6 \varepsilon_0}=\frac{ q }{2 \varepsilon_0}$

$n =3$

Alternatively, $\phi \propto$ no of electric field lines passing through surface $q$ is point charge which has uniformly distributed electric field lines thus half of electric field lines will pass through hemisphere $\&$ other half will pass through conical surface.