Average force on an electron in a wire

Hey, just did a 25 question multiple choice paper and this was the only one that I got wrong. Was hoping someone could enlighten me.

Quote:

Originally Posted by **AQA PHYA4/1 Jan 2010**

A current of $\displaystyle 8.0\, A$ is passed through a conductor of length $\displaystyle 0.40 \,m$ and cross-sectional area $\displaystyle 1.0\times 10^{-6} \:m^2$. The conductor contains $\displaystyle 8.0\times 10^{28}$ electrons per $\displaystyle m^3$. When the conductor is at right angles to a magnetic field of flux density $\displaystyle 0.20\, T$, it experiences a magnetic force. What is the average magnetic force that acts on one of the free electrons in the wire?

I got the force acting on the conductor:

$\displaystyle

F = BIl = 0.20 \times 8.0 \times 0.40 = 0.64N $

So figured since that is the force on the entire conductor, then the force on one electron would be the force divided by the number of electrons:

$\displaystyle

\frac{F}{N_e} = \frac{0.64}{8.0 \times 10^{28}} = 8.0 \times 10^{-30} \, N $

But the answer is $\displaystyle 2.0 \times 10^{-23} \, N $

I know I am neglecting the cross-sectional area but not sure where it would come in.

Thanks.