Millimoles of $H_{2}S{O}_4=20\times 0.1=2$

Millimoles of $N{H}_{4}OH=30\times 0.2=6$



Hence, the solution will act as a basic buffer.

$pOH=p{K}_b+\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{t}\right]}{\left[\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{e}\right]}$

$\left[N{H}_4^{+}\right]=2\times \left[{\left(N{H}_4\right)}_{2}S{O}_4\right];{\left[N{H}_4\right]}^{+}=4$

$pOH=p{K}_b+\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[N{\mathrm{H}}_4^{+}\right]}{\left[N{H}_3\right]}$

$=4.7+\mathrm{l}\mathrm{o}\mathrm{g}\frac{4}{2}$

$=4.7+0.3010=5.010$

$pH=14-pOH=9$