Answer
a) ${\bf 9.4\times 10^{15}}\;\rm proton$
b) ${\bf 115}\;\rm A/m^2$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the current is given by
$$I=\dfrac{Q}{t}=\dfrac{Ne}{t}$$
where $Q=Ne$ where $e$ is the charge of the proton.
So the number of protons delivered by the beam in one second is given by
$$N=\dfrac{tI}{e}$$
Plug the known;
$$N=\dfrac{(1)(1.5\times 10^{-3})}{(1.6\times 10^{-19})}$$
$$N=\color{red}{\bf 9.375\times 10^{15}}\;\rm proton$$
$$\color{blue}{\bf [b]}$$
We know that the current density is given by
$$J=\dfrac{I}{A}$$
Hence,
$$dI=dA J$$
The dashed circle, in the figure below, represents a hollow cylinder or a thin ring that has a radius of $r$ and a thickness of $dr$. Hence, its cross-sectional area is given by $dA=2\pi r dr$
$$dI=2\pi r dr J$$
where $J=J_{\rm edge}\dfrac{r}{R}$
$$dI=\dfrac{ 2\pi J_{\rm edge}}{R}r^2 dr $$
Integrating both sides;
$$\int_0^IdI=\dfrac{ 2\pi J_{\rm edge}}{R}\int_0^Rr^2 dr $$
$$I=\dfrac{ 2\pi J_{\rm edge}}{3R} r^3\bigg|_0^R $$
$$I=\dfrac{ 2\pi J_{\rm edge}}{3R} R^3 =\frac{ 2 }{3 } \pi J_{\rm edge}R^2$$
Solving for $J_{\rm edge}$;
$$J_{\rm edge}=\dfrac{3I}{2\pi R^2 }$$
Plug the known;
$$J_{\rm edge}=\dfrac{3(1.5\times 10^{-3})}{2\pi (2.5\times 10^{-3})^2 }$$
$$J_{\rm edge}=\color{red}{\bf 114.6}\;\rm A/m^2$$
