Just put your origin somewhere on the cylinder's axis, which we choose as the z axis. The problem is translation invariant along this axis and under rotations around this axis. So you introduce cylinder coordinates (r,\varphi,z. I guess that the whole question is about electrostatics. Then the electric field has a potential \vec{E}=-\vec{\nabla} \Phi, and due to the symmtries this potential depends only on r.
Then Gauß's Law tells you
\vec{\nabla} \cdot \vec{E}=-\Delta \Phi=\rho. \qquad (1)
The charge density is
\rho=\sigma \delta(\rho-b).
The Poisson equation (1) reads for our case
\frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r} \left (r \frac{\mathrm{d} \Phi}{\mathrm{d} r} \right )=-\rho.
Now we have nearly everywhere \rho=0. So we solve the differential equation for \rho=0. We simply can multiply by r and integrate ones. This gives
r \frac{\mathrm{d} \Phi}{\mathrm{d} r}=C_1.
Dividing by r, and integrating again gives
\Phi(r)=C_1 \ln \left (\frac{\rho}{\rho_0} \right ). \qquad(2)
Here, I have written the second integration constant such as to make the log's argument dimensionless as it must be.
Now we need to work out the boundary conditions. First of all in the conducting inner cylinder we must have \vec{E}=0 and thus \Phi=\text{const}. For convenience we set \Phi(r)=0 there. Now there is no charge on this inner cylinder, and the potential must be smooth on the boundary r=a. Thus we have
\Phi(r)=0 \quad \text{for} \quad r \leq b.
Now we must work in the surface-charge density at r=b, where according to (1) the potential must be continuous but its first derivative must jump such that the \delta distributions singularity is correctly described. Multiplying (1) with r and integrating wrt. r between r=b-0^+ and r=b+0^+ yields
b \left [\Phi'(b+0^+)-\Phi'(b-0^+) \right ]=-\sigma b.
Since \Phi'(b-0^+)=0, we must have
\Phi'(b+0^+)=-\sigma.
Taking the derivative of our solution (2), we find
\frac{C_1}{b}=-\sigma \; \Rightarrow \; C_1=-\sigma b.
The constant \rho_0 must be chosen such that \Phi(b)=0, i.e., \rho_0=b. Thus our solution for the potential reads
\Phi(r)=\begin{cases}<br />
0 & \text{for} \quad r \leq b, \\<br />
-\sigma b \ln \left (\frac{r}{b} \right ) & \text{for} \quad r>b.<br />
\end{cases}<br />
The electric field is
\vec{E}=-\vec{e}_r \Phi'(r)=\begin{cases}<br />
0 & \quad \text{for} \quad r<b \\<br />
\sigma b/r \vec{e}_r & \quad \text{for} \quad r>b.<br />
\end{cases}<br />
Note, that the normal component of \vec{E} which is E_r makes the correct jump at r=b, namely the surface-charge sensity \sigma. So we have found the complete solution of the problem.
