The current flowing clockwise in the equilateral triangle has a magnetic field in the direction $\widehat k$.

$\tau $ = BiNAsin$\theta $ = BiAsin90°

$ \therefore $ A = ${\tau \over {iB}}$

Area of equilateral triangle

A = $${1 \over 2}\left( l \right)\sqrt {{l^2} - {{\left( {{l \over 2}} \right)}^2}} $$ = $${{\sqrt 3 } \over 4}{l^2}$$

$ \Rightarrow $ $${{\sqrt 3 } \over 4}{l^2} = {\tau \over {iB}}$$

$ \Rightarrow $ $$l = 2{\left( {{\tau \over {\sqrt 3 Bi}}} \right)^{{1 \over 2}}}$$