If you integrate $\rho$ over the volume, the $\delta$ will be zero everywhere except where its argument is $0$. Since the area under this function is normalized to $1$, the result of the integration will be that the charge will be $0$ everywhere except at $r=0$ where it will be $q$.
This follows because you want $\rho(r)$ to be $0$ everywhere except at $r=0$ (hence the $\delta(r)$), and you want $\int dV \rho(r)=q$ since this is the only charge in your system, hence the $q\delta(r)$.