To solve this problem it is necessary to apply the concepts related to Current and Load.
The current in terms of the charge of an electron can be expressed as

Where,
q = Charge
t = time
At the same time the Charge is the amount of electrons multiplied by the amount of these, that is
q = ne
Replacing in the first equation we have to


Clearing n,

Here the time is one second then



Therefore the number of electrons per second are passing any cross sectional area of the wire are
