Given the inner cylindrical surface of radius "a" is charged with a constant uniform charge density of $\rho_s$, How did they derive this formula for Flux Density at distance r radially:
$\vec{D} = \rho_s \frac{a}{r}\hat{r}~~~[\frac{C}{m^2}]$
Where:
$\rho_s = \frac{\text{Charge}}{\text{Area}}~~~[\frac{C}{m^2}]$
$\vec{D} \propto \frac{1}{r}$
The symmetry ensure us that, at all points between the two surfaces, $\vec D$ is oriented as $\hat r$. and the Gauss Theorem , in your notation, say that, for the the flux, we have:
$\Phi=2\pi l r D= Q= \rho_s 2\pi a l $
From this we have: $ D=\frac{a}{r}\rho_s $ and the vector $\vec D$ has the direction $\hat r$ :
$\vec D=\frac{a}{r}\rho_s \hat r$