(1) Number density.

First find the proton speed from its kinetic energy. Convert energy to joules:

$$E_k = 2 \times 10^{7} \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.2 \times 10^{-12} \text{ J}$$

Using $E_k = \frac{1}{2} m v^2$ (non-relativistic approximation):

$$v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-12}}{1.7 \times 10^{-27}}} \approx 6.14 \times 10^{7} \text{ m/s}$$

The beam current relates to number density by $I = n A v e$, where $A = \pi r^2$ is the cross-section and $e$ is the proton charge. Solving:

$$n = \frac{I}{A v e}$$

With $A = \pi (1.0 \times 10^{-3})^2 \approx 3.14 \times 10^{-6} \text{ m}^2$:

$$n = \frac{1.0 \times 10^{-3}}{3.14 \times 10^{-6} \times 6.14 \times 10^{7} \times 1.6 \times 10^{-19}} \approx 3.24 \times 10^{13} \text{ m}^{-3}$$

(2) Number of protons in 1 min.

The total charge delivered in time $t$ is $Q = I t$, and the number of protons is:

$$N = \frac{I t}{e} = \frac{1.0 \times 10^{-3} \times 60}{1.6 \times 10^{-19}} \approx 3.75 \times 10^{17}$$