Answer
a) $1.125\times 10^4 \;\rm ion$
b) $2.55\times 10^7 \;\rm A/m^2$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that a current of 1.8 pA is carried by a potassium ion.
So,
$$I=\dfrac{Q}{t}$$
So the total charge amount of potassium ion is given by
$$Q=Ne=tI$$
where $Q=Ne$ where $N$ is the number of ions and $e$ is the charge of one ion.
Thus, the number of ions that pass during 1 ms is given by
$$N=\dfrac{tI}{e}$$
Plug the known;
$$N=\dfrac{(1\times 10^{-3})(1.8\times 10^{-12})}{(1.6\times 10^{-19})}$$
$$N=\color{red}{\bf 1.125\times 10^4}\;\rm ion$$
$$\color{blue}{\bf [b]}$$
The current density is given by
$$J=\dfrac{I}{A}=\dfrac{I}{\pi r^2}$$
Plug the known;
$$J= \dfrac{(1.8\times 10^{-12})}{\pi (0.15\times 10^{-9})^2}$$
$$J=\color{red}{\bf 2.55\times 10^7}\;\rm A/m^2$$
