An arrangement of three parallel straight wires placed...

Force per unit length between two parallel long current carrying wires is $= \dfrac{{\text{\mu}}_{\text{0}} {I}_{1} {I}_{2}}{\text{2 \pi d}}$
Here, force between BC and AB will be same in magnitude.

${F}_{BC} = {F}_{BA} = \dfrac{{\mu}_{0}{I}^{2}}{2\pid}$

${F}_{res} = \sqrt{{F}_{BC}^{2} + {F}_{BA}^{2}}$

$⟹ {F}_{res} = \dfrac{\sqrt{2}{\mu}_{0}{I}^{2}}{2\pid} = \dfrac{\mu{}_{0}{I}^{2}}{\sqrt{2}\pid}$