A particle is moving with constant speed in a circular path. When...

In circular motion with constant speed, magnitude of instantaneous velocity, ${v}_{ins} = \omegar$.

Time taken to turn by $90 °$ will be $= \dfrac{\dfrac{\pi}{2}}{\omega} = \dfrac{\pi}{2\omega}$.

Displacement will be equal to the length of $AB = \sqrt{{r}^{2} + {r}^{2}} = r\sqrt{2}$

Average velocity

${v}_{avg} = \dfrac{\text{displacement}}{\text{time}} = \dfrac{r\sqrt{2} \times 2\omega}{\pi} = \dfrac{2\sqrt{2}}{\pi}\omegar$

Hence, required ratio $\dfrac{{v}_{\text{ins}}}{{v}_{\text{avg}}} = \dfrac{\omegar\pi}{2\sqrt{2}\omegar} = \dfrac{\pi}{2\sqrt{2}}$

Therefore, $x = 2$.