Write an expression for the volume charge density $\rho(\mathbf{r})$ of a point charge $q$ at $\mathbf{r}'$ in terms of the Dirac delta function.
My attempt:
Let $\mathbf{r}-\mathbf{r'}=\mathbf{R}$.
$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}\hat{\mathbf R}$$
Taking the divergence of both sides, \begin{align} \nabla\cdot\mathbf{E}(\mathbf{r}) & =\frac{q}{4\pi\epsilon_0}\nabla\cdot \left(\frac{\hat{\mathbf{R}}}{R^2}\right) \\ \implies\frac{\rho(\mathbf{r})}{\epsilon_0} & =\frac{q}{4\pi\epsilon_0} 4\pi\delta^3(\mathbf{R}) \\ \implies\rho(\mathbf{r}) & =q\delta^3(\mathbf{r}-\mathbf{r}') \tag{1} \end{align}
However, the dimensions of the left hand side are 'charge per volume' whereas the dimensions of the right hand side are just 'charge' as $\delta^3(\mathbf{r}-\mathbf{r}')$ is dimensionless. Where have I gone wrong?