c
Initially $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{s}}$
Finally
$\mathrm{B}_{2}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{1}^{2}+2 \mathrm{B}_{1} \mathrm{B}_{2} \cos 120^{\circ}}$
$=2 \mathrm{B}_{1} \cos 60^{\circ}=\mathrm{B}_{1}$
So, $B$ remains unchanged.
