The speed of a small object undergoing uniform circular motion is $4m/s.$ The magnitude of the change in the velocity during 0.5 seconds is also $4m/s.$

If the angular displacement is $\theta$ for 0.5 seconds, then the magnitude of change in velocity is $\Delta V=\sqrt{v^2+v^2-2v^2\cos \theta }=2v\sin \frac{\theta }{2}$

$\therefore \Delta V=2\times 4\sin \frac{\theta }{2}$ Or $\frac{\theta }{2}=\frac{\pi }{6}\&\theta =\frac{\pi }{3}$

$\therefore \omega =\frac{\theta }{0.5}=\frac{2\pi }{3}rad/s$

$R=\frac{v}{\omega }=\frac{4\times 3}{2\pi }=\frac{6}{\pi }m$ and centripetal acceleration $=v\times \omega \times \frac{2\pi }{3}=\frac{8\pi }{3}m/s^2.$