When the proton is rotated in a cyclotron its maximum velocity will be corresponding to the maximum radius of the circular path. We know that the radius of the circle in a magnetic field is given by,

$r=\frac{mv}{Bq}\Rightarrow v=\frac{Bqr}{m}$.

The kinetic energy will be,

$$\begin{gathered} K=\frac{1}{2}m{v}^2=\frac{B^2{q}^2{r}^2}{2m} \\ \Rightarrow K=\frac{1^2\times {\left(1.6\times {10}^{-19}\right)}^2\times {\left(0.6\right)}^2}{2\times 1.6\times {10}^{-27}\times 1.6\times {10}^{-13}} \mathrm{MeV} \\ \Rightarrow K=18 \mathrm{MeV} \end{gathered}$$