Answer
(a) $\lambda =2 \pi \sigma R$
(b) $E = \dfrac{\sigma R}{\epsilon_{0} r}$
(c) $E = \dfrac{\lambda}{2 \pi \epsilon_{0} r} $
Work Step by Step
(a) For the cylinder shape, the charge from the linear charge density is the same for the charge calculated from the surface charge density
\begin{gather*}
q = q\\
\lambda l=\sigma A\\
\lambda l=\sigma 2 \pi R l\\
\boxed{\lambda =2 \pi \sigma R}\\
\end{gather*}
(b) We could get the electric field from the expression of Gauss's law for the electric flux by
\begin{gather*}
\Phi_{E}=\frac{q}{\epsilon_{0}}\\
2 \pi r l E=\frac{\sigma(2 \pi R l)}{\epsilon_{0}}\\
\boxed{E = \frac{\sigma R}{\epsilon_{0} r}} \tag{1}
\end{gather*}
(c) Plug the expression of $\sigma$ into equation (1) to get the electric field in terms of $\lambda$
$$ E=\frac{\sigma R}{\epsilon_{0} r}=\frac{R}{\epsilon_{0} r}\left(\frac{\lambda}{2 \pi R}\right)=\boxed{\frac{\lambda}{2 \pi \epsilon_{0} r} }$$
