Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that the colume charge density is
$$\rho =\dfrac{C}{r^2}\tag 1$$
where $C$ is constant.
To solve this problem, we need to use the hint given by the author.
$$dq=\rho dV$$
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),
$$dq=\dfrac{C}{r^2} (4\pi r^2) dr$$
$$dq=4\pi C dr\tag 2$$
Integrating;
$$\int_0^Qdq=\int_0^R4\pi C dr$$
$$Q=4\pi C R$$
Hence,
$$\boxed{ C=\dfrac{Q}{4\pi R}}$$
$$\color{blue}{\bf [b]}$$
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),
$$Q_{in}=\int_0^{r} 4\pi C dr=4\pi Cr$$
Plug $C$ from the boxed formula,
$$Q_{in} =\dfrac{Q}{4\pi R}4\pi r=\dfrac{Qr}{R}$$
Plug into (3),
$$ E =\dfrac{Qr}{4\pi r^2 \epsilon_0R} $$
$$ E =\dfrac{Q }{4\pi \epsilon_0Rr} $$
Therefore,
$$ \boxed{E =\dfrac{1 }{4\pi \epsilon_0}\dfrac{Q}{Rr}\;\hat r} $$
$$\color{blue}{\bf [c]}$$
when $r=R$,
$$E =\dfrac{1 }{4\pi \epsilon_0}\dfrac{Q}{R^2}$$
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 $r\geq R$.
