Let's play with resistance

So, you can make any system of flowing charge from one to another and conductivity will be alway the same. The only thing you should care is that your simulation will allow you to calculate $G$ without unknown terms.

Okey then, lets give $q$ to one and $-q$ to another sphere. Because they are placed on a much bigger distance compared to their radius their potential will be

$\phi_1=\frac{1}{4\pi\epsilon_0} \frac{q}{r}$

$\phi_2=\frac{1}{4\pi\epsilon_0} \frac{q}{r}$

And then voltage is

$U=\frac{1}{2\pi\epsilon_0} \frac{q}{r}$ .

The electric field on the surface of the sphere is given by formula

$E=\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$

Using Ohm's law one can easy get that

$J=\sigma E$

You should notice that all charge left one sphere will end up on the other so the net current is $I=J(r) 4\pi r^2$ .

and now using give formula you will find that:

$G=2\pi\sigma r$

Correction:

Actually the only one small part of charge won't get to second sphere (red field line) that one goes to infinity but it's very small compared to net charge so we don't think about it.