b
The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat{ k }$
$\tau=\operatorname{BiN} A \sin \theta$
$\tau=\operatorname{BiN} A \sin 90^{\circ}$
$\tau= Bi \times \frac{\sqrt{3}}{4} I ^{2} \times 1$
$\left[\because\right.$ Area of equilateral triangle $=\frac{\sqrt{3}}{4} I ^{2}$ and $\left. N =1\right]$
$\Rightarrow I ^{2}=\frac{4 \tau}{\sqrt{3} Bi } \Rightarrow I =2\left[\frac{\tau}{ Bi \sqrt{3}}\right]^{1 / 2}$