First things first, you are going to have to rename one of your variables. The r in the \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} is not the same r as the radius of the cylindrical shell. You'll have to rename one of them to avoid confusion. [The problem statement tells you to change one of them to d.]
Second, can you use Gauss' Law to derive the \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} formula with only the wire (ignoring the cylindrical shell for the moment)? If you know how to do that, adding in the cylindrical shell is pretty straightforward.
Let's look at Gauss's Law:
\oint \vec E \cdot \vec {dA} = \frac{Q_{enc}}{\varepsilon_0}
The dA is a differential area element on your chosen Gaussian surface. Choose your Gaussian surface wisely to take advantage of symmetry. If you choose it wisely in certain special cases like this one, the closed integral is easy to solve. The idea is that you want the dot product of the electric field and the differential area element to be constant over the entire surface [Edit: possibly ignoring things like cylindrical endcaps, if you can show that \vec E \cdot \vec{dA} = 0 at the endcaps]. That way the left hand side simply becomes EA, where A represents a formula for the Gaussian surface's area. (You can't do this for everything, but you can do this for a) things with spherical symmetry, b) infinitely long wires/cylinders [like this problem] or c) infinite planes.)
The right hand side only depends on the charge enclosed
within the Gaussian surface. Anything outside the Gaussian surface doesn't matter.
Then solve for E. You can add the unit direction vector in as a final step.
