Explanation

(d) Since, the currents in the three wires are flowing in same direction so, the wire $B$ will experience a force of attraction due to both wires $A\ and\ C,$
So, $F_{AB} = \frac{\mu_{0}}{4\pi}\ .\frac{2i_{A}i_{B}}{d} = \ \frac{\mu_{0}}{4\pi}\ .\frac{2 \times 1 \times 2}{d}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4\mu_{0}}{4\pi d}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
and $F_{CB} = \ \frac{\mu_{0}}{4\pi}\ .\frac{2i_{B}i_{C}}{d} = \frac{\mu_{0}}{4\pi} \times \frac{2 \times 2 \times 3}{d}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{12\mu_{0}}{4\pi d}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
As seen from Eqs. (i) and (ii) $F_{CB} > F_{AB}$ hence, the net force of attraction will be directed towards wire $C.$