Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that
$$E=E_{max}\left( \dfrac{r^4}{R^4}\right)$$
So, $E=E_{max}$ when $ \dfrac{r^4}{R^4}=1$.
In other words, $E=E_{max}$ when $r=R$ and hence in this case,
$$\boxed{E_{max}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R^2}}$$
$$\color{blue}{\bf [b]}$$
We can use Gauss's law when $r\lt R$,
$$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
Substitute $E$ from the given
$$E_{max}\left( \dfrac{r^4}{R^4}\right)(4\pi r^2)=\dfrac{Q_{in}}{\epsilon_0}$$
Substitute $E_{max}$ from the boxed formula above,
$$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R^2}\left( \dfrac{r^4}{R^4}\right)(4\pi r^2)=\dfrac{Q_{in}}{\epsilon_0}$$
$$ \dfrac{Q}{\epsilon_0 }\left( \dfrac{r^6}{R^6}\right) =\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$Q_{in}=Q\left( \dfrac{r^6}{R^6}\right) \tag 1$$
Now we need to find $Q_{in}$ in terms of $\rho$,
$$dq=\rho dV$$
where$\rho$ is the volume charge density and is a function in $r$, $dV$ is the volume of a spherical shell of radius $r$, and thickness $dr$. Hence, $dV=(4\pi r^2) dr$.
Hence,
$$dq=\rho (4\pi r^2) dr$$
Integrating;
$$\int_0^{Q_{in}}dq=\int \rho (4\pi r^2) dr$$
$$Q_{in}=4\pi \int \rho r^2 dr$$
Substituting from (1),
$$Q\left( \dfrac{r^6}{R^6}\right) =4\pi \int \rho r^2 dr$$
$$Q\left( \dfrac{r^6}{4\pi R^6}\right) =\int \rho r^2 dr$$
Taking the first derivative for both sides of $d/dr$
$$Q\left( \dfrac{6r^5}{4\pi R^6}\right) =\dfrac{d}{dr}\left[ \int \rho r^2 dr\right]$$
$$Q\left( \dfrac{3r^5}{2\pi R^6}\right) = \rho r^2 $$
Therefore,
$$\boxed{\rho= \dfrac{3Qr^3}{2\pi R^6} }$$
$$\color{blue}{\bf [c]}$$
Using the same approach we used above for a thin spherical shell of radius $r$, thickness $dr$,
$$dq=\rho dV=\rho (4\pi r^2 dr)$$
Integrating, when $r\geq R$
$$\int_0^{Q_{in}}dq = 4\pi \int_0^R\rho r^2 dr $$
substituting $\rho$ from the boxed formula above,
$$Q_{in} = 4\pi \int_0^R \dfrac{3Qr^3}{2\pi R^6} r^2 dr $$
$$Q_{in} = \dfrac{ 6Q}{ R^6} \int_0^R r^5 dr $$
$$Q_{in} = \dfrac{ 6Q}{ R^6} \dfrac{ r^6 }{6}\bigg|_0^R$$
$$Q_{in} = \dfrac{ QR^6}{ R^6} =Q$$
which is the total charge of the sphere.
