66If charges +q and &ndash q are placed on the two spheres, then the potential difference between then is
V=\dfrac{2q}{4\pi\epsilon_0 R}
R being the radius.
The field on the surface of each sphere is
E=\dfrac{q}{4\pi\epsilon_0 R^2}
Hence the current density (which is radial) is
j=\sigma E=\dfrac{\sigma q}{4\pi\epsilon_0 R^2}
with σ being the conductivity. So the total current leaving one sphere (and entering the other one) is
I =j \,4\pi R^2 =\dfrac{\sigma q}{\epsilon_0 }
Hence, the total resistance between the two conductors is
r=\dfrac{V}{I}=\dfrac{1}{2\pi\sigma R}=\dfrac{\rho}{2\pi R}
Plug in the values to get x=\pi ohms