A copper wire of cross-sectional area 2.0 ##mm^2## carries a current of 10 A. How many electrons pass through a given cross-section of the wire in one second ?

##6.3 * 10^19"e"^(-)"s"^(-1)##

This is a classic example of a trick question. Sort of.

The last thing you need to worry about is the cross-section of the wire. Here's why.

What is an ampere?

An ampere is the equivalent of one coulomb per second. In your case, a current of ##"10 A"## is equivalent to

##"10 A" = "10 coulomb"/"s"##

How about a coulomb?

A coulomb is the equivalent of roughly ##6.241 * 10^(18)## elementary charges, or

##"1 C" = 6.241 * 10^(18) xx underbrace(1.60217662 * 10^(-19)"C")_(color(blue)("elementary charge"))##

So if a total charge of ##"10 C"## is passing through the cross-section per second, how many electrons would that be equivalent to?

Well, if ##"1 C"## is equivalent to ##6.241 * 10^(18)## electrons per second, ##"10 C"## will be equivalent to

##10 color(red)(cancel(color(black)("C"))) xx (6.241 * 10^(18)e^(-)"s"^(-1))/(1color(red)(cancel(color(black)("C")))) = 6.241 * 10^(18)"e"^(-)"s"^(-1)##

Now, the elemental charge is often given as ##1.6 * 10^(-19)"C"##, which means that you would get ##6.25 * 10^(18)## electrons in one coulomb.

In that case, the answer will indeed be

##10 color(red)(cancel(color(black)("C"))) xx (6.25 * 10^(18)e^(-)"s"^(-1))/(1color(red)(cancel(color(black)("C")))) = 6.3 * 10^(18)"e"^(-)"s"^(-1)##

Rounded to two , of course.

So remember, think about the basic concepts and don't get distracted by "additional information", which can sometimes be misleading.