Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the potential difference through an inductor is given by
$$\Delta V_L=-L\dfrac{dI}{dt}$$
where $I=I_0e^{-t/\tau}$, so
$$\Delta V_L=-I_0L\dfrac{d }{dt}e^{-t/\tau}$$
$$\boxed{\Delta V_L=\dfrac{I_0L}{\tau}e^{-t/\tau}}$$
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above,
$$ \Delta V_L=\dfrac{(50\times 10^{-3})(20\times 10^{-3})}{(1\times 10^{-3})}e^{-t/(1\times 10^{-3})}$$
$$ \Delta V_L= e^{-t/(1\times 10^{-3})}$$
$\bullet\Rightarrow$ At $t=0$ s,
$$ \Delta V_L= e^{-0/(1\times 10^{-3})}=\color{red}{\bf 1}\;\rm V$$
$\bullet\Rightarrow$ At $t=1$ ms,
$$ \Delta V_L= e^{-(1\times 10^{-3})/(1\times 10^{-3})}=\color{red}{\bf 0.368}\;\rm V$$
$\bullet\Rightarrow$ At $t=2$ ms,
$$ \Delta V_L= e^{-(2\times 10^{-3})/(1\times 10^{-3})}=\color{red}{\bf 0.135}\;\rm V$$
$\bullet\Rightarrow$ At $t=3$ ms,
$$ \Delta V_L= e^{-(3\times 10^{-3})/(1\times 10^{-3})}=\color{red}{\bf 0.05}\;\rm V$$
