A cyclotron has dees of radius 0.40 m in a field of 1.6 T…

Step 1. The frequency. $f = \dfrac{qB}{2\pi m} = \dfrac{3.2 \times 10^{-19}\ \text{C} \times 1.6\ \text{T}}{2\pi \times 6.68 \times 10^{-27}\ \text{kg}} = 1.22 \times 10^{7}\ \text{Hz}$, that is $12.2$ MHz.

Step 2. The final energy. First $qBR = 3.2 \times 10^{-19}\ \text{C} \times 1.6\ \text{T} \times 0.40\ \text{m} = 2.05 \times 10^{-19}\ \text{kg m/s}$. Then $K_{\max} = \dfrac{(qBR)^2}{2m} = \dfrac{(2.05 \times 10^{-19}\ \text{kg m/s})^2}{2 \times 6.68 \times 10^{-27}\ \text{kg}} = 3.14 \times 10^{-12}\ \text{J}$. In MeV that is $\dfrac{3.14 \times 10^{-12}}{1.6 \times 10^{-13}} = 19.6$ MeV.

Step 3. The turns. Each crossing gives $qV = 3.2 \times 10^{-19}\ \text{C} \times 2.5 \times 10^{4}\ \text{V} = 8.0 \times 10^{-15}\ \text{J}$. The number of crossings is $\dfrac{3.14 \times 10^{-12}}{8.0 \times 10^{-15}} = 392$. Two crossings make one turn, so there are $196$ complete turns.