The electric field of earth is really quite complicated. This paper is old, but presents a nice overview of the issues. That $500\,000\, \mathrm{C}$ number is related a charge separation between the surface and the bottom of the ionosphere, rather than the net charge on the earth. And in any case, it is only the result of a rough model whose underpinnings aren't entirely accurate. Your small sphere is embedded inside this massive system (the Earth's atmosphere). As @John Rennie points out in the comments, because the Earth is so large, you can ignore its curvature, and a better model for this system is a small sphere between the two plates of an enormous capacitor with a $\sim 300\,000\, \mathrm{V}$ potential difference.
So, as always, when dealing with potentials, you need to ask yourself: potential relative to what? Well, assuming that the ionosphere is actually at the same potential as a point infinitely far away, the natural interpretation of your question is to assume that your sphere is neutral infinitely far away, is insulated, transported to the surface of the earth, and then touched to the surface of the earth, and ask how much charge is transferred at this last step. My calculations suggest that a $1\,\mathrm{m}$ sphere has a capacitance of about $100\, \mathrm{pF}$. For a voltage difference of $\sim 300\,000\, \mathrm{V}$, that would be a few times $10^{-5}\,\mathrm{C}$.
Incidentally, as my answer on this question suggests, humans are modeled as capacitors with about the same capacitance, so you would get the same amount of charge.
Of course, these calculations are all for isolated objects, which makes them not perfectly valid, but should be reasonable approximations.
The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is:
$$ F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2} $$
and the field strength at the surface of the second sphere is:
$$ F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2} $$
Lets take the ratio $F_1/F_2$ to see which is greater. The constants cancel to give us:
$$ \frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}} $$
and I'm going to rewrite this slightly to make it obvious how you use your equality $Q_1/r_1 = Q_2/r_2$:
$$ \frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}} $$
Because $Q_1/r_1 = Q_2/r_2$ we can cancel them on the top and bottom of the fraction and we're left with:
$$ \frac{F_1}{F_2} = \frac{r_2}{r_1} $$
and because $r_2 < r_1$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1.
Best Answer
The potential at the surface of a charged sphere or radius $r$ is:
$$ V = k\frac{Q}{r} \tag{1} $$
Since the area of the sphere is $4\pi r^2$, the charge density is:
$$ \rho = \frac{Q}{4\pi r^2} $$
and rearranging gives:
$$ Q = \rho 4\pi r^2 $$
and substituting for $Q$ in equation (1) we get:$$ V = 4\pi k\rho r $$
Can you take it from here?