Explanation

(d) $6\Omega$ and $6\Omega$ are in series, so effective resistance is $12\Omega$ which is in parallel with $3\Omega,$ so
$\frac{1}{R} = \frac{1}{3} + \frac{1}{12} = \frac{15}{36}$
$\Rightarrow R = \frac{36}{15}$
$\therefore I = \frac{V}{R} = \frac{4.8 \times 15}{36} = 2A$