d
${B}=3\left[\frac{\mu_{0} {i}}{4 \pi {r}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)\right]$
$\tan 60^{\circ}=\frac{\ell / 2}{{r}}$
Where ${r}=\frac{9 \times 10^{-2}}{2 \sqrt{3}} \,{M}$
$\therefore {B}=3 \times 10^{-5}\, {T}$
Current is flowing in clockwise direction so, $\overrightarrow{{B}}$ is inside plane of triangle by right hand rule.
