Answer
a) ${\bf 4.55\times 10^{-13}}\;\rm J $
b) ${\bf 2844}\;\rm rev$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The maximum kinetic energy of the proton occurs when it is at maximum velocity which occurs when the proton reaches the edge diameter of the cyclotron.
$$K_{\rm max}=\frac{1}{2}mv_{\rm max}^2\tag 1$$
Now we need to find $v_{\rm max}$. We know that the only force exerted on the proton inside a cyclotron is the magnetic force which is given by
$$F_B=qvB\sin90^\circ=ma_r$$
where $a_r$ is the radial acceleration which is given by $v^2/R$, so
$$qvB=\dfrac{mv^2}{R}$$
$$q B=\dfrac{mv }{R}$$
$$q B=\dfrac{mv_{\rm max} }{R_{\rm max}}$$
Hence,
$$v_{\rm max} =\dfrac{q BR_{\rm max}}{m}$$
Plug into (1),
$$K_{\rm max}=\frac{1}{2}m\left[\dfrac{q BR_{\rm max}}{m}\right]^2$$
$$K_{\rm max}= \dfrac{q^2 B^2R^2_{\rm max}}{2m} $$
Plug the known;
$$K_{\rm max}= \dfrac{(1.6\times 10^{-19})^2 (0.75)^2 (32.5\times 10^{-2})^2 }{2(1.67\times 10^{-27})} $$
$$K_{\rm max}= \color{red}{\bf 4.55\times 10^{-13}}\;\rm J $$
$$\color{blue}{\bf [b]}$$
To find the number of revolutions, we need to find the energy gained during one full cycle and then divide that by the maximum kinetic energy gained.
The energy gained in on full cycle is given by
$$U_{\rm cycle}=2q\Delta V$$
the 2 is because that the proton accelerates through the potential difference of 500 V two times during one full cycle.
Hence,
$$N=\dfrac{K_{\rm max}}{U_{\rm cycle}}=\dfrac{K_{\rm max}}{2q\Delta V}$$
Plug the known;
$$N= \dfrac{4.55\times 10^{-13} }{2(1.6\times 10^{-19})(500)}$$
$$N=\color{red}{\bf 2844}\;\rm rev$$
