A current of i amp is flowing in an equilateral triangle of side l The magnetic induction at the centroid will be

See fig  The magnetic induction at O due to current i in arm AB is given by 
${\mathrm{B}}_{\mathrm{AB}}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi }(\mathrm{l}/2\sqrt{3})}\left(\sin {60}^{\circ }+\sin {60}^{\circ }\right)=\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}$
similarly ${\mathrm{B}}_{\mathrm{BC}}=\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}\text{ and }{\mathrm{B}}_{\mathrm{AC}}=\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}$
$\therefore {\mathrm{B}}_{\mathrm{ABC}}=\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}+\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}+\frac{3{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}=\frac{9{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}$