If there were no space between nuclei, how big would the Earth be? I have some people telling me it would be the size of baseball. I am quite doubtful on this. If this is true, then the gaps must be so incredibly huge that everything should be transparent. I am not a physics student, just wondering.
 A: Instead of assuming the earth is made of metallic hydrogen, let's just compare Earth's density of $5.52 \times 10^3 kg/m^3$ to that of neutrons' $2.3 \times 10^{17} kg/m^3$ because degenerate matter consisting of neutrons is what you get when electrons are forced into nuclei. That's a density increase of about $4.17 \times 10^{13}$ (at least 3 orders of magnitude different from Pranav's proton-atom volume ratio estimate), or a radial decrease of about $3.47 \times 10^4$. 
This puts a neutron Earth radius at 184 metres or a huge baseball STADIUM including car park. 
A: When you look at crystalline substances, there is really not that much space between the atoms. What people mean when they say that an atom is mostly empty space, is that the INSIDE of the atom is very sparsely populated with stuff.
This is because the stuff in question, the nucleus and the electrons, are tiny in comparison to the actual size of the atom. The nucleus is typically a few hundred thousandths of an angstrom unit, or $\approx 10^{-15} \text{meters}$ in diameter. The entire atom, on the other hand, has a diameter to the order of an angstrom, or $10^5$ times larger than the nucleus. The electrons are even tinier, so much so that the volume that each electron occupies is negligible in comparison to the nucleus.
A simple calculation will tell you that the volume occupied by an atom is $\approx 10^{15}$ times the volume of its building blocks, for the Hydrogen atom.
A similar calculation for an atom of a heavier element would tell you $V_{atom} \approx 10^{10} \times V_{nucleus}$
If you applied this reduction, by removing all that empty volume inside the atom, to the Earth, you would get an 
$$
V_{earth}'=V_{earth} \times 10^{-12} \\
d_{earth}' = d_{earth} \times 10^{-4} = 1300 \text{m}
$$
P.S.: The empty space isn't really empty space. The currently accepted model of the atom says that an electron can be anywhere inside that space, but has a higher probability of being in a particular part of it. As David pointed out, it is more accurate to say "Increasing the density of the earth to the density of the nucleus, while keeping the mass the same."
A: Suppose that the Earth could be compressed and turned into a neutron star. Neutron stars have overall densities up to $5.9\times 10^{17}\,\text{kg/m}^3$ (source). Since the total mass of the Earth is about $5.9\times 10^{24}\,\text{kg}$, it would be a sphere with a volume of about $10^7\,\text{m}^3$, which corresponds with a radius of about $134\,\text{m}$.
A: earth volume is about $1,083,206,916,846$ cubic kilometers. The hydrogen atom is about $99.9999999999996\%$ empty. Based on this, if you take the empty space out of earth it becomes as big as a swimming pool of $1000\times 100 \times 1$ meters.
A: 0.8768 femtometers. the same size as a proton.
