Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that the colume charge density is
$$\rho =\rho_0\left[ 1-\dfrac{r}{R}\right] \tag 1$$
where $\rho_0$ is constant.
Now we need to solve for $\rho_0$,
$$dq=\rho dV$$
where $dV$ is the volume of a spherical shell of radius $r$, and thickness $dr$. Hence, $dV=(4\pi r^2) dr$.
Hence,
$$dq=\rho (4\pi r^2) dr$$
Plug $\rho $ from (1),
$$dq=\rho_0\left[ 1-\dfrac{r}{R}\right] (4\pi r^2) dr \tag 2$$
Integrating;
$$\int_0^Qdq=\int_0^R\rho_0\left[ 1-\dfrac{r}{R}\right] (4\pi r^2) dr $$
$$Q=4\pi\rho_0\int_0^R \left[ r^2-\dfrac{r^3}{R}\right] dr $$
$$Q=4\pi\rho_0 \left[ \dfrac{r^3}{3}-\dfrac{r^4}{4R}\right]_0^R $$
$$Q=4\pi\rho_0 \left[ \dfrac{R^3}{3}-\dfrac{R^4}{4R}\right] $$
$$Q=\dfrac{\pi\rho_0R^3}{3} $$
$$\boxed{\rho_0=\dfrac{3Q}{\pi R^3} }$$
$$\color{blue}{\bf [b]}$$
when $r\lt R$,
$$\oint \vec E\cdot d\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
$$ E A =\dfrac{Q_{in}}{\epsilon_0}$$
$$ E =\dfrac{Q_{in}}{A \epsilon_0}$$
$$ E =\dfrac{Q_{in}}{4\pi r^2\epsilon_0}\tag 3$$
Now we need to find $Q_{in}$,
$$Q_{in}=\int dq$$
Plug from (2),
$$Q_{in}=\int_0^{r} \rho_0\left[ 1-\dfrac{r}{R}\right] (4\pi r^2) dr $$
$$Q_{in}=4\pi \rho_0\int_0^{r} \left[ r^2-\dfrac{r^3}{R}\right] dr $$
$$Q_{in}=4\pi\rho_0 \left[ \dfrac{r^3}{3}-\dfrac{r^4}{4R}\right]_0^r $$
$$Q_{in}=4\pi\rho_0 \left[ \dfrac{r^3}{3}-\dfrac{r^4}{4R}\right] $$
$$Q_{in}=4\pi\rho_0 \left[ \dfrac{4Rr^3-3r^4}{12R}\right] $$
Plug $\rho_0$ from the boxed formula,
$$Q_{in}=4\pi\dfrac{3Q}{\pi R^3} \left[ \dfrac{4Rr^3-3r^4}{12R}\right] $$
$$Q_{in}= \dfrac{ Q}{ R^3} \left[ \dfrac{4Rr^3-3r^4}{R}\right] $$
Plug into (3),
$$ E =\dfrac{1}{4\pi r^2\epsilon_0} \dfrac{ Q}{ R^3} \left[ \dfrac{4Rr^3-3r^4}{R}\right] $$
$$ E =\dfrac{Q}{(4\pi \epsilon_0)R^3} \left[ {4 r -\dfrac{3r^2}{R}} \right] $$
$$ E =\dfrac{Qr}{(4\pi \epsilon_0)R^3} \left[ {4 -\dfrac{3r }{R}} \right] $$
Therefore,
$$ \boxed{E =\dfrac{Qr}{(4\pi \epsilon_0)R^3} \left[ {4 -\dfrac{3r }{R}} \right]\;\hat r} $$
the direction is outward since the magnitude of the electric field is positive since $r\lt R$.
$$\color{blue}{\bf [c]}$$
when $r=R$,
$$ E =\dfrac{QR}{(4\pi \epsilon_0)R^3} \left[ {4 -\dfrac{3R }{R}} \right] =\dfrac{Q }{(4\pi \epsilon_0)R^2} $$
which is a reasonable result since this is the electric field of a point charge at distance $R$ since the spherical charge behaves as if the entire charge were at the center when $r\geq R$.
