a
Here, side of the triangle, \(l=4.5 \times 10^{-2} \,\mathrm{m}\) current, \(I =1 \,\mathrm{A}\) 

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 

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