Perhaps I may make the following approximation? The voltage is turned on so slowly that the electrons find their way close to the surface in an ordered manner and then slowly expose themselves on the surface. This way the current density is the same over the whole surface and only in the surface-normal direction.
I think “surface current” is somewhat misleading. I would think of a current along the surface. This will happen when you have the wire connected to one side of the capacitor and let the charges flow in. Here I will assume that there the change in voltage is so slow that the tangential currents can be neglected and we only want the current that is normal to the surface.
The charge $Q$ that a capacitor can hold on one of the plates depends on the capacity $C$ and voltage $V$ like $Q = CV$. The new charges that get to the surface when the voltage is increased by $\Delta V$ is $\Delta Q = C \, \Delta V$. These new charges are evenly distributed on the whole surface.
For a current $I$ we also have $Q = I T$, after a time $T$ the charge $Q$ will be transferred by the current. A current density $i$ can be obtained by dividing through the area $A$, so $i = I/A$.
From this you should be able to compute the current density at the surface. If you are stuck or finished, you can continue reading.
We divide the above relation by some short time interval $\Delta t$, take the limit $\Delta t \to 0$ and end up with a time derivative. So the expression is $\dot Q = C \dot V$. But $\dot Q = I$ already. We divide by the surface area $A$ of the capacitor and obtain $i = C \dot V / A$.