Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The induced emf for a wire that moves in a magnetic field is given by
$$\varepsilon=vlB$$
Plug the known and referring to Figure 33.26.
$$\varepsilon=(100)(0.04)(1.0)=\color{red}{\bf 4}\;\rm V$$
$$\color{blue}{\bf [b]}$$
The induced current is given by
$$I=\dfrac{\varepsilon}{R}$$
where $R_{net}=4R=4(0.01)=0.04\;\Omega$;
$$I=\dfrac{4}{0.04}=\color{red}{\bf 100}\;\rm A$$
$$\color{blue}{\bf [c]}$$
The moving wire works as a battery. It has an internal resistance of 0.01$\Omega$, so the potential difference between its ends is then
$$\Delta V=\varepsilon-IR_{\rm wire}=4-(100)(0.01)$$
$$\Delta V=\color{red}{\bf 3}\;\rm V$$