Solution:

Since $$\tau =I {\frac {d\omega} {dt}}= mB\, sin\, \theta$$ $$\ldots\left(i\right)$$

NOW, $$\frac{d \omega}{dt}=\frac{d\omega}{d\theta} \frac{d\theta}{dt}$$

$$=\omega \frac{d\omega}{d\theta} \ldots\left(ii\right)$$

then, from $$\left(i\right)$$ and$$ \left(ii\right)$$

$$I\omega \frac{d\omega}{d\theta}=mB\,sin\,\theta ; I\omega d\omega=mBsin\theta d\theta$$

Integrating both sides

$$I \int\limits_{0}^{\omega_{f}} \omega\, d\omega=mB \int\limits_{0}^{45^{\circ}} sin\, \theta\,d\theta$$

$$I \left[ \frac{\omega^{2}}{2}\right]_{0}^{\omega_{f}} =mB\left[-cos \,\theta\right]_{0^{\circ}}^{45^{\circ}}$$

$$\frac{1}{2}I \omega_{f}^{2}=-mB\left[cos\,45^{\circ}-cos\,0^{\circ}\right]$$

$$=-mB\left[\frac{1}{\sqrt{2}}-1\right]=0.29\, mB$$

$$\omega_{f}^{2}=\frac{2\times0.29mB}{f}$$

$$=\frac{2\times0.29\times15\times4}{0.50}$$

$$\omega_{f}^{2}=69.6 ; \omega_{f} =\sqrt{69.6}$$

$$=8.34\, rad \, s^{-1}$$solution