Answer
a) $120\;\rm C$
b) $0.45\;\rm mm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the current is given by
$$I=\dfrac{Q}{t}$$
So the charge amount is given by
$$Q=It$$
Plug the known;
$$Q=(150)(0.8)$$
$$Q=\color{red}{\bf 120}\;\rm C$$
$$\color{blue}{\bf [b]}$$
To find the distance traveled by one electron on the wire during the 0.8-s, we need to find its drift speed.
$$v_{\rm d}=\dfrac{J}{n_ee}=\dfrac{I}{An_ee}$$
where $v=s/t$, so
$$v_{\rm d}=\dfrac{s}{t}=\dfrac{I}{An_ee}$$
Hence, the distance traveled by one electron is given by
$$s= \dfrac{It}{An_ee}= \dfrac{It}{\pi r^2e (n_e)_{\rm copper}}$$
Plug the known;
$$s= \dfrac{(150)(0.8)}{\pi (2.5\times 10^{-3})^2(1.6\times 10^{-19}) (8.5\times 10^{28})}$$
$$s=\color{red}{\bf 0.45}\;\rm mm$$
