The angular velocity corresponds to the radians that the radius sweeps in a second. Let's introduce an intermediate quantity: the period $T$. That is the time it takes for the wheel to complete one spin.
If we know that the wheel takes $T$ seconds to make a complete rotation, then calculating the angular velocity $\omega$ is easy: the angle swept in a whole rotation is $2\pi$ radians. Then we have
$$\omega = \frac{2\pi}{T}\quad\text{[rad/s] or [$\mathrm s^{-1}$]}$$
In your problem there's a different quantity: the frequency. It is measured in $\mathrm s^{-1}$ and indicates how many whole rotations occur in one second. From the problem we know that $f = 200/60\ \mathrm s^{-1}$. You can verify that between frequency and period there exists the relation $T = f^{-1}$. Therefore we obtain another formula for $\omega$:
$$\omega = 2\pi f\quad\text{[rad/s] or [$\mathrm s^{-1}$]}$$
Finally, tangential velocity is directly proportional to the angular velocity, and the constant of proportionality is the radius, i.e.
$$v = \omega r\quad\text{[$\mathrm m\mathrm s^{-1}$]}$$