From Gauss law,

${\mathrm{\varphi }}_{\mathrm{hemisphere}\mathrm{ }}+{\mathrm{\varphi }}_{\mathrm{Cone}\mathrm{ }}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

Total flux produced from q in α angle

$\mathrm{\varphi }=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}[1-\mathrm{cos\alpha }]$ 

For hemisphere, α = $\frac{\mathrm{\pi }}{2}$

${\mathrm{\varphi }}_{\mathrm{hemisphere}\mathrm{ }}=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}$ 

From equation (i)

$=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}+{\mathrm{\varphi }}_{\mathrm{cone}\mathrm{ }}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ 

${\mathrm{\varphi }}_{\mathrm{cone}\mathrm{ }}=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}$ 

$\frac{4\mathrm{q}}{6{\mathrm{\epsilon }}_0}=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}$ 

n = 3

Alternatively, φ ∝ 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.