magnetic field at the centre of the triangle $'O'B = ?$

From figure, $\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}$

$\Rightarrow d=\frac{l}{2 \sqrt{3}}=\left(\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}\right)\, \mathrm{m}$

Magnetic field, $B=\frac{\mu_{0} i}{4 \pi d}\left(\cos \theta_{1}+\cos \theta_{2}\right)$

Putting value of $\mu=4 \pi \times 10^{-7}$ and $\theta_{1}$ and $\theta_{2}$

we will get net magenetic field

$=3 \times B=4 \times 10^{-5} \,\mathrm{Wb} / \mathrm{m}^{2}$