1. Write the reaction

The neutralization is:

$H_2SO_4 + 2NH_4OH \rightarrow (NH_4)_2SO_4 + 2H_2O$

Since $H_2SO_4$ is dibasic, $1$ mole of $H_2SO_4$ neutralizes $2$ moles of $NH_4OH$.

2. Calculate moles of reactants

For sulfuric acid:

$n(H_2SO_4)=0.1\times 0.020=0.002\text{ mol}$

This provides acidic equivalents:

$2\times 0.002=0.004\text{ mol of }H^+$

For ammonium hydroxide:

$n(NH_4OH)=0.2\times 0.030=0.006\text{ mol}$

3. Find excess reactant after neutralization

$0.004$ mol of base is consumed by the acid, so leftover base:

$0.006-0.004=0.002\text{ mol}$

Salt formed:

$0.002\text{ mol }H_2SO_4 \Rightarrow 0.002\text{ mol }(NH_4)_2SO_4$

Each mole of $(NH_4)_2SO_4$ gives $2$ moles of $NH_4^+$, so ammonium ion formed:

$2\times 0.002=0.004\text{ mol }NH_4^+$

Thus the final mixture contains a basic buffer of weak base $NH_4OH$ and its conjugate acid $NH_4^+$.

4. Use buffer formula for basic buffer

For a basic buffer:

$\text{pOH}=pK_b+\log\frac{[salt]}{[base]}$

Here,

$pK_b=4.7$

and

$\frac{[salt]}{[base]}=\frac{0.004}{0.002}=2$

So,

$\text{pOH}=4.7+\log 2$

Using $\log 2\approx 0.3$,

$\text{pOH}=4.7+0.3=5.0$

Hence,

$\text{pH}=14-5.0=9.0$

5. Check options

The resultant pH is:

$\boxed{9.0}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.