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I guess the important thing to realise is that the size atoms is irrelevant to the problem. If we considered two atoms joined together to be a new "atom" then they would be twice as heavy, so the forces would be four times as strong, but there would be only half as many atoms, so there would be four times fewer pairs.
So the answer is just the integral as r and r' range over the interior of the earth of G ρ(r) ρ(r')/(r-r')^2, where ρ(r) is the density. We can assume constant density, but I still can't be bothered to do the integral.
The earth has mass 5.97*10^24 kg and radius 6.37*10^6 m, G = 6.674*10^-11 m^3 kg^-1 s^-2 and we want an answer in Newtons = m kg s^-2. So by dimensional analysis, the answer is about G M^2/r^2 = 5.86*10^25.
That doesn't seem right, though? Imagine a one dimensional version of the problem. If a stick of length 1 is divided into n atoms weighing 1/n each, then each pair of adjacent atoms is distance 1/n apart, so the force between them is 1. Since there are n such pairs, the total force grows at least linearly with n. And it gets even worse if some atoms are disproportionately closer to others (in molecules).