Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The easiest way to solve this problem is to use the Gaussian surfaces.
Assume that there is a Gaussian sphere inside the sphere of radius $r$ where $r\lt R$.
Now we need to find $\Delta V$ between the center of the sphere $V_0$ and a point at $r$ distance from this center and inside the sphere $V_r$.
$$\Delta V=V_r-V_0=-\int E_{\rm in}dr$$
Plug from the given $E_{\rm in}$ formula,
$$ V_r-V_0=-\int_0^r \dfrac{rQ}{4\pi \epsilon_0 R^3}dr$$
$$ V_r =V_0- \dfrac{r^2Q}{8\pi \epsilon_0 R^3} \tag 1$$
Now we need to find $V_0$ where we know that at the surface of the sphere, the potential is considered outside the sphere and is then given by
$$V_R=\dfrac{Q}{4\pi \epsilon_0R}\tag 2$$
Using (1),
$$V_R-V_0=-\dfrac{R^2Q}{8\pi \epsilon_0 R^3}$$
$$V_0=V_R+\dfrac{ Q}{8\pi \epsilon_0 R }$$
Plug from (2),
$$V_0=\dfrac{Q}{4\pi \epsilon_0R}+\dfrac{ Q}{8\pi \epsilon_0 R }$$
$$V_0= \dfrac{ 3Q}{8\pi \epsilon_0 R }\tag 3$$
Plug into (1),
$$ V_r =\dfrac{ 3Q}{8\pi \epsilon_0 R }- \dfrac{r^2Q}{8\pi \epsilon_0 R^3} $$
$$ \boxed{V_r =\dfrac{ Q}{8\pi \epsilon_0 R }\left[3- \dfrac{r^2 }{ R^2} \right] }\tag 4$$
$$\color{blue}{\bf [b]}$$
The ratio is given by
$$\dfrac{V_{\rm center}}{V_{\rm surface}}=\dfrac{V_0}{V_R}$$
Plug from (2) and (3);
$$\dfrac{V_{\rm center}}{V_{\rm surface}}=\dfrac{ \dfrac{ 3Q}{8\pi \epsilon_0 R }}{\dfrac{Q}{4\pi \epsilon_0R}}=\color{red}{\bf \dfrac{3}{2}} $$
$$\color{blue}{\bf [c]}$$
At outside the sphere when $r=3R$, using (4),
$$ V_{3R}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{Q}{3R}$$
$$ V_{R}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{Q}{R}$$
$$ V_{0}=\dfrac{ 1}{4\pi \epsilon_0 }\dfrac{3Q}{2R}$$
See the graph below,