(d)
Millieq. of 0.01 M $HCl = \frac{0.01 \times 100}{1000} = 1 \times 10^{- 3}$
$\therefore pH = 3$
Millieq. of 0.02 M $H_{2}SO_{4} = \frac{0.04 \times 50}{1000} = 2 \times 10^{- 3}$
Millieq. of 0.02 M $NaOH = \frac{0.02 \times 50}{1000} = 1 \times 10^{- 3}$
Left $\left\lbrack H^{+} \right\rbrack = 2 \times 10^{- 3} - 1 \times 10^{- 3};\ \therefore pH = 3$