Q:

Gauss's Law in action

Need someone to tell me if I got this done correctly
(a) Draw Gaussuian cylinder inside the black cylinder to find charge enclosed
$Q_{en} = Q(\frac{r}{a})^2$
Apply Gauss's Law
$E2\pi r \ell = \dfrac{Q\left ( \frac{r}{a} \right )^2}{\varepsilon_0}$
$E = \dfrac{Q\left ( \frac{r}{a} \right )^2}{2 \pi r \ell\varepsilon_0} = \dfrac{Qr}{2\pi \varepsilon_0  a^2\ell}$
$E = \dfrac{\rho r}{2 \varepsilon_0}$
(b) Basically same principle, but the total charge of the black cylinder 
$E = \dfrac{\rho a^2}{2r \varepsilon_0}$
(c) 0 under static equilibrium
(d) The enclosed charge should be a sum of cylinders' charges. 
$Q_{en} = \rho V + \lambda \ell = \rho \pi a^2 \ell + \lambda \ell = \ell ( \rho \pi a^2 + \lambda)$
Hence
$E2\pi r \ell = \dfrac{\ell ( \rho \pi a^2 + \lambda)}{\varepsilon_0} \implies E =\dfrac{ \rho \pi a^2 + \lambda}{2 \pi r \varepsilon_0} $
The term looks very awkward, so it's probably wrong. 
Thank you for reading

A:

As for a), note that $Q=\rho \pi a^2 l$.

