Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
After a long time
$$\boxed{\Delta V_C=\varepsilon}$$
$$\color{blue}{\bf [b]}$$
The charge in the capacitor is given by
$$Q=C\Delta V_C$$
and since after a long time $\Delta V_C=\varepsilon$, then
$$\boxed{Q_{\rm max}=C\varepsilon}$$
$$\color{blue}{\bf [c]}$$
since the capacitor is charging, which means that the charge is increasing inside the capacitor,
$$I=\dfrac{+dQ}{dt}$$
$$\color{blue}{\bf [d]}$$
The current here decreases with time until it becomes zero when $\Delta V_C=\varepsilon$, so
$$I=\dfrac{dQ}{dt}=\dfrac{d }{dt}\left[ Q_{\rm max}-Q_{\rm max}\;e^{-t/RC}\right]$$
$$I =Q_{\rm max}\;\dfrac{d }{dt}\left[ 1-e^{-t/RC}\right]$$
$$I =Q_{\rm max}\; \left[ -\dfrac{-1}{RC}e^{-t/RC}\right]$$
Plug $Q_{\rm max}$ from above,
$$I = \dfrac{ C\varepsilon}{RC}e^{-t/RC} $$
$$\boxed{I = \dfrac{ \varepsilon}{R }e^{-t/RC} }$$
