The magnetic field at the centre O due to the current through side AB is

$B_1=\frac{{\mu }_{0}I}{4\pi a}\left[\sin {\theta }_1+\sin {\theta }_2\right]$



As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field at O is sum of all the fields.

i.e. $B=3B_1=\frac{3{\mu }_{0}I}{4\pi \theta }$ $\left[\sin {\theta }_1+\sin {\theta }_2\right]$

Here, $\tan {\theta }_1=\frac{AD}{OD}$

$\Rightarrow \tan {60}^o=\frac{\frac{1}{2}}{a}$

$\Rightarrow a=\frac{1}{2\sqrt{3}}=\frac{9\times {10}^{-2}}{2\sqrt{3}}$

Now $B=3\times \frac{4\pi \times {10}^{-7}\times 2}{4\pi \times \frac{9\times {10}^{-2}}{2\sqrt{3}}}$ $[\sin {60}^o+\sin {60}^o]$

$=\frac{4\sqrt{3}}{9}\times {10}^{-6}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)$

$=\frac{4}{3}\times {10}^{-6}$

$\therefore B=1.33\times {10}^{-6} \mathrm{T}$