What you are missing is how to change the sum into an integral.
Divide the material into $n$ sheets parallel to the plates, of infinitesimal thickness $\delta=\frac{L}{n}$ and area $A$, where $L$ is the distance between the plates. The resistance of each sheet is $\frac{\rho \delta}{A}=\frac{\delta}{\sigma A}$. The sheets are in series so the resistance of the whole slab is the sum of each :
$R=\frac{\delta}{\sigma_1 A}+\frac{\delta}{(\sigma_1+d\sigma)A}+\frac{\delta}{(\sigma_1+2d\sigma)A}+$ ... $+\frac{\delta}{(\sigma_1+(n-1)d\sigma)A}$
where $d\sigma=\frac{\sigma_2-\sigma_1}{n}$ is the increment of conductivity between sheets .
This infinite sum can be converted into an integral :
$R=\Sigma_0^{\infty} \frac{\delta(n)}{\sigma(n) A}=\frac{L}{(\sigma_2 -\sigma_1)A} \int_{\sigma_1}^{\sigma_2} \frac{d\sigma}{\sigma}=\frac{L}{(\sigma_2 - \sigma_1)A} \ln (\frac{\sigma_2}{\sigma_1})$.
The rest of the question is just a matter of applying Ohm's Law.
