Considering the annular region of the cylindrical capacitor as the Gaussian surface, we have,
$E\times A=\frac{Q}{\epsilon }$. The surface area is given by,
$A=2\pi r\times L$, $\Rightarrow E\times 2\pi r\times L=\frac{Q}{\epsilon }$. 
Considering $Q’$as the charge per unit length, we get,
$Q\text{'}=\frac{Q}{L}$$\Rightarrow E=\frac{Q\text{'}}{2\pi r\epsilon }\Rightarrow E\propto \frac{1}{r}$.