Answer
${\bf 20}\;\rm V$
Work Step by Step
Recalling, from the previous chapters, that the current through a discharged capacitor is given by
$$I=I_{\rm max}e^{-t/\tau}$$
where $_{\rm max}$ is the maximum current which is given by
$$I_{\rm max}=\dfrac{(\Delta V_C)_{\rm max}}{R}$$
where $\tau=RC$, so $R=\tau/C$;
$$I_{\rm max}=\dfrac{C(\Delta V_C)_{\rm max}}{\tau }$$
Solving for $(\Delta V_C)_{\rm max}$;
$$(\Delta V_C)_{\rm max}=\dfrac{\tau I_{\rm max}}{C}$$
From the given formula, we can see that $I_{\rm max}=10$ A, $\tau=2\;\mu$s.
Plug the known into the previous formula,
$$(\Delta V_C)_{\rm max}=\dfrac{ (2\times 10^{-6} )(10)}{(1\times 10^{-6})}$$
$$(\Delta V_C)_{\rm max}=\color{red}{\bf 20}\;\rm V$$
