Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that the colume charge density is
$$\rho =\dfrac{r\rho_0}{R}$$
where $\rho_0$ and $R$ are constants.
This means that this a linear function.
$$\rho =\dfrac{\rho_0}{R}r\tag 1$$
Noting that this works when $r\leq R$, so when $r\gt R$, $\rho=0$
See the first figure below.
$$\color{blue}{\bf [b]}$$
To solve this problem, we need to use the hint given by the author.
$$dq=\rho dV$$
where $dV$ is the volume of a cylinderical shell of length $L$, radius $r$, and thickness $dr$. Hence, $dV=(2\pi r)L dr$.
Hence,
$$dq=2 \pi \rho rL dr$$
Plug $\rho $ from (1),
$$dq= \dfrac{2 \pi \rho_0 L}{R} r^2 dr$$
Integrating;
$$\int_0^Q dq=\int_0^R \dfrac{2 \pi \rho_0 L}{R} r^2 dr$$
$$Q=\dfrac{2 \pi \rho_0 L}{3R} r^3 \bigg|_0^R $$
$$Q=\dfrac{2 \pi \rho_0 L}{3R} (R^3-0) $$
$$Q=\frac{2}{3 } \pi \rho_0 LR^2 $$
where $Q=\lambda L$
$$\lambda \color{red}{\bf\not} L=\frac{2}{3 } \pi \rho_0 \color{red}{\bf\not} LR^2 $$
Hence,
$$\boxed{\rho_0=\dfrac{3\lambda}{2\pi R^2} }$$
$$\color{blue}{\bf [c]}$$
when $r\lt R$,
$$\oint \vec E\cdot d\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
$$\int_{\rm top} \vec E\cdot d\vec A+\int_{\rm bottom} \vec E\cdot d\vec A+\int_{\rm sides} \vec E\cdot d\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
$$ 0+0+E A_{\rm sides} =\dfrac{Q_{in}}{\epsilon_0}$$
$$ E =\dfrac{Q_{in}}{A_{\rm sides}\epsilon_0}$$
$$ E =\dfrac{Q_{in}}{2\pi r L\epsilon_0}\tag 2$$
Now we need to find $Q_{in}$,
$$Q_{in}=\int_0^{Q_{in}} dq=\int_0^r \dfrac{2 \pi \rho_0 L}{R} r^2 dr$$
$$Q_{in} = \dfrac{2 \pi \rho_0 L}{3R} r^3 $$
Plug into (2),
$$ E =\dfrac{1}{2\pi r L\epsilon_0}\dfrac{2 \pi \rho_0 L}{3R} r^3 $$
$$ E = \dfrac{ \rho_0 }{3\epsilon_0 R} r^2 $$
Plug $\rho$ from the boxed formula in part (b),
$$ E = \dfrac{ 1 }{3 \epsilon_0R}\dfrac{3\lambda}{2\pi R^2} r^2 $$
Therefore,
$$\boxed{ E = \dfrac{ \lambda r^2}{2\pi \epsilon_0R^3}\;\hat r} $$
$$\color{blue}{\bf [d]}$$
when $r=R$,
$$E = \dfrac{ \lambda R^2}{2\pi \epsilon_0R^3} = \dfrac{ \lambda }{2\pi \epsilon_0R } $$
$$E = \dfrac{1}{4\pi \epsilon_0}\dfrac{ 2\lambda }{ R } $$
which is a reasonable result since this is the electric fiel of an infinite line of charge.
