The centre of a wheel rolling on a plane surface moves with a speed $v_0$. A particle on the rim of the wheel at the same level as the centre will be moving at speed

As per question,

The speed of the centre of wheel $=v_o$

Particle is on the rim of the wheel,

So,

The situation can be shown as in the diagram
Here $v_0=R\omega$
At $P,v=r\omega$
$=\sqrt{(R^2+R^2)\omega }=\sqrt{2}R\omega =\sqrt{2}{v}_0$

So the speed of the particle is on the rim $=\sqrt{2}R\omega$