Answer
${\bf 20.4}\;\rm V$
Work Step by Step
To find the voltage of the capacitor, we need to find the charge inside it.
We know that $I=dQ/dt$, so
$$Q=\int_{0\;\rm ms}^{\rm 20\; ms} Idt$$
where $I=I_{\rm max}e^{-t/\tau}$
$$Q=I_{max}\int_{0\;\rm ms}^{\rm 20\; ms} e^{-t/\tau} dt$$
$$Q=I_{max}\left[ -\tau e^{-t/\tau}\right]_{0\;\rm ms}^{\rm 20\; ms}$$
$$Q=-\tau I_{max}\left[ e^{-t/\tau}\right]_{0\;\rm ms}^{\rm 20\; ms}$$
$$Q=65 \left[-40 e^{-20/40} +40\right] =\bf 1.02\;\rm mC$$
Now we can find the voltage of the capacitor at $t=20$ ms,
$$\Delta V_C=\dfrac{Q}{C}=\dfrac{1.02\times 10^{-3}}{50\times 10^{-6}}=\color{red}{\bf 20.4}\;\rm V$$
