$pH$ of 0.01 M $(NH_4)_2SO_4$ and 0.02 M $NH_4OH$ buffer ($pK_a$ of $NH_4^+$ = 9.26) is

(a) 4.74 + log 2

(b) 4.74 - log 2

(c) 9.26 + log 2

(d) 9.26 + log 1

$pK_a(NH^+_4) = 9.26$
$pK_b(NH_3)$ = 14 - 9.26 = 4.74

$[NH^+_4] = 2 \times [(NH_4)_2SO_4] = 2 \times 0.01 = 0.02 M$

$pOH = pK_b + log \frac{[NH^+_4]}{[NH_4OH]}$
$\Rightarrow pOH = 4.74 + log \frac{0.02}{0.02}$ = 4.74 - log 1

pH = 14 - pOH = 14 - 4.74 + log 1 = 9.26 + log 1