Answer
a) ${\bf 2.5}\;\rm C$
b) ${\bf 1.8}\;\rm cm$
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 given;
$$Q=(50\times 10^3)(50\times 10^{-6})$$
$$Q=\color{red} {\bf 2.5}\;\rm C$$
$$\color{blue}{\bf [b]}$$
We know, according to Ohm's law, that
$$\Delta V=IR$$
where $R=\rho L/A$ and $A=\pi r^2=\pi D^2/4$ where $D$ is the diameter.
$$\Delta V_{max}=I\dfrac{4\rho_{\rm iron} L}{\pi D_{min}^2}$$
Solving for $D_{min}$;
$$D_{min} = \sqrt{\dfrac{4\rho_{\rm iron} I L}{\pi \Delta V_{max}}}$$
Plug the known;
$$D_{min}=\sqrt{\dfrac{4(9.7\times 10^{-8})(50\times 10^3)(5)}{\pi (100)}}$$
$$D_{min}=\color{red}{\bf 1.8}\;\rm cm$$
