Given, the angular velocity, $\omega =\frac{2\pi }{60}\times 120 \mathrm{rad} {\mathrm{s}}^{-1}$. 

Recall the formula of induced EMF in terms of angular velocity and radius of the ring, $\epsilon =\frac{1}{2}B\omega r^2$, now put the proper values at proper places, 

$\Rightarrow \epsilon =\frac{1}{2}\times \left(0.4\times {10}^{-4}\right)\times \left(2\pi \left[\frac{120}{60}\right]\right)(1{)}^2$

$\Rightarrow \epsilon =0.8\pi \times {10}^{-4}$$=2.512\times {10}^{-4} \text{V}$.