As shown in the figure, a current of 2A flowing in an equilateral...

Perpendicular distance of centroid from any of the sides will be, $dtan60 ° = 2\sqrt{3}$

$⇒ d = 2cm$

Each side will subtend an angle of $120 °$ at the centre. Also magnetic field due to all sides will be out of plane of paper, hence total magnetic field can be written as

$B = 3 \times \dfrac{{\mu}_{0}i}{4\pi d}\left[\sin 60 ° + \sin 60 °\right]$

$= 3 \times \dfrac{{10}^{- 7} \times 2}{2 \times {10}^{- 2}} \times \sqrt{3}$

$= 3\sqrt{3} \times {10}^{- 5}T$