Well... the number of electrons in the balls depends on what isotope they consist of.
If they are made solely of the most common iron isotope, Fe-56, then yes, there are
26 protons à 1.0072765 u,
30 neutrons à 1.0086650 u and
26 electrons à 5.4857990x10^-4 u
per one atom.
This kind of an atom, however, does not simply weigh just as much as the components put together. Together, the components weigh less than the sum of their masses, because some energy is stored into forces that keep the nucleus together. So you can't just sum the neutrons and protons together, because a nucleus formed by them weigh less than that. You have to either know the average binding energy of individual nucleods, or then you can check the literature value for Fe-56 isotope's atomic mass, which is...
M (Fe-56) = 55,85 u
When if you simply sum the particles together, you would end up presuming that the mas sper atom would be ~56.463 u.
I think there's the slight mistake you made.
Just calculate again, with slightly less mass per atom, and you end up with more atoms and more electrons removed, and thus you increase the result for force, and it should go right.
EDIT: Actually the bigger error is this:
in 5.6g, or .0056kg, there should be .0056/9.384e-26 = 5.96737e22 atoms of iron.
one in a billion of that is 5.96737e13, wich is the number of removed electrons.
It's not, because there is 26 electrons per atom.
Mass per atom (presuming it's all Fe-56) is 55.85 u = 9.274117017x10^-26 kg.
Mass of a single ball is 5.6x10^-3 kg.
Atoms per single ball are
5.6x10^-3 / 9.274117017x10^-26 = 6.0383106981x10^22
Amount of electrons is 26 times that amount, 'cause there's 26 electrons per atom, thus...
1.56996078153x10^24 electrons in total/ball.
Of those, one billionth are removed. That's
1.56996078153x10^24 / 10^9 = 1.56996078153x10^15 electrons per ball removed.
So, each ball is charged at 1.56996078153x10^15 e.
e = 1.6021773x10^-19 Coulombs, so the charge per ball is
[EDIT2: Fixored line] Q = 2.51535552606x10^-4 C
From which you can easily calculate the force. Which is outwards from each ball. Any good?
I'm actually getting 227457.39794 N as a result. but that's in the right magnitude all right, so it's probably just a matter of how it's calculated, how accurately things like mass loss in nuclei and the isotope configuration are taken into account.