66Okay, here is the solution. Usually, when encountering problems having infinite extensions, superposition principle comes to mind. Here also, we use superposition to combine separate discussions of current flowing in and out. First, we send the current into A and collect it at infinity. Then, we send the current at infinity and collect it at B. When we superpose the two situation, we get the situation of the problem.
When the current I0 is sent into A and collected at infinity, it is distributed hemispherically symmetrically in the half plane containing the conducting medium. That means that the current density j at a distance r from A is given by
j(r)=\dfrac{I_0}{2\pi r^2}
As per Ohm's law in the local form \vec{j}(r)=\dfrac{\vec{E}(r)}{\rho} (here \rho is the resistivity and \vec{E}), we get the magnitude of electric field at a distance r from A as
E(r)=\dfrac{I_0\rho}{2\pi r^2}
The potential at a distance r can be obtained by integrating the field. However, in this case, we get it by comparison of a point charge. We note that the field of a point charge varies as \dfrac{1}{r^2} and the potential only as \dfrac{1}{r} and the coefficient of \dfrac{1}{r^2} and \dfrac{1}{r} are the same. In the present case also the field is varying as \dfrac{1}{r^2}, so the potential must be
V(r)=\dfrac{I_0\rho}{2\pi r}
Hence, the potential difference between C and D, when current is sent at A and collected at infinity is
\Delta V_1=\dfrac{I_0\rho}{2\pi a}-\dfrac{I_0\rho}{2\pi a\sqrt{2}}=\dfrac{I_0\rho}{4\pi a}(2-\sqrt{2})
We next send the current at infinity and collect it as B. Everything is the same except the signs of the quantities. The potential difference between C and D is therefore the same.
So when we superpose, the potential difference between C and D becomes twice of what we calculated above. Hence,
V_0=2\Delta V_1=\dfrac{\rho I_0}{2\pi a}(2-\sqrt{2})
Accordingly, the resistivity can be found as
\boxed{\rho = (2+\sqrt{2})\dfrac{\pi a V_0}{I_0}}