Answer
$-{ \rho R^2}/{4\epsilon_0}$
Work Step by Step
The easiest way to solve this problem is to use the Gaussian surfaces.
Assume that there is a Gaussian cylinder inside the cylinder of radius $r$ where $r\lt R$. And the length of this Gaussian cylinder is $L$.
We are given that $\rho $ is the volume charge density, so
$$\rho=\dfrac{Q_{\rm enclosed}}{V_{\rm Gaussian \;cylinder }}$$
Hence,
$$Q_{\rm enclosed}=\rho V_{\rm Gaussian \;cylinder }$$
$$Q_{\rm enclosed}=\rho \pi r^2 L\tag 1$$
According to Gauss's law,
$$\oint EdA=E(2\pi r L)$$
Hence,
$$ 2\pi r LE=\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
Plug from (1),
$$ 2 \color{red}{\bf\not}\pi \color{red}{\bf\not}r \color{red}{\bf\not}LE=\dfrac{\rho \color{red}{\bf\not} \pi r^{ \color{red}{\bf\not}2} \color{red}{\bf\not} L}{\epsilon_0}$$
$$E=\dfrac{ r \rho}{2\epsilon_0}\tag 2$$
Recalling that $\Delta V=-\int Eds$,
So,
$$\Delta V=-\int_0^REdr$$
Plug from (2),
$$\Delta V=-\dfrac{ \rho}{2\epsilon_0}\int_0^Rrdr$$
$$\Delta V=-\dfrac{ \rho}{4\epsilon_0} r^2\bigg|_0^R$$
$$\boxed{\Delta V=-\dfrac{ \rho R^2}{4\epsilon_0} }$$
