Explanation

(d) Effective value of resistance of parallel combination of $20\Omega,\ 30\ \Omega,\ 60\Omega$ is $R_{1},\ where$
$\frac{1}{R_{1}} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3 + 2 + 1}{60} = \frac{6}{60} = \frac{1}{10}$
$R_{1} = 10\Omega$
Similarly effective value of parallel combination of $24\Omega$ and $8\Omega$ resistance is given by
$R_{2} = \frac{24 \times 8}{24 + 8} = 6\Omega$
Hence, the circuit may be redrawn as shown in the adjacent figure, where total resistance across $A$ and $B$, $R = 3 + 10 + 6 + 1 = 20\Omega$. As potential across $R_{2}( = 66\Omega)$ is 48 V, hence
$V_{AB} = 48 \times \frac{R}{R_{2}} = \frac{48 \times 20}{6} = 160V$