The given data is 

$v=\pi \mathrm{m} {\mathrm{s}}^{-1}$

By using trigonometric ratios in the triangle shown above the displacement can be found out by

$$\begin{gathered} \sin 60°=\frac{AB}{2R} \\ \Rightarrow AB=R\sqrt{3} \end{gathered}$$

The formula for distance is 

$$\begin{gathered} d=vt \\ \Rightarrow t=\frac{d}{v} \\ \Rightarrow t=\frac{120°}{360°}\times \frac{2\pi R}{\pi }=\frac{2R}{3} \end{gathered}$$

So, the average velocity is 

$v_{\mathrm{avg}}=\frac{\sqrt{3}R}{{\displaystyle \frac{2R}{3}}}=\frac{3\sqrt{3}}{2}$

$=1.5\sqrt{3} \mathrm{m} {\mathrm{s}}^{-1}$