Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).
However, the assumption that $g_{00}$ is constant implies that \begin{equation} \label{em2l} A=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As result scalar Ricci curvature is constant, too ($R=-6~b$). Your metric describes homogeneous, static and isotropic solution, the so-called Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .