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 (Einstein field equations) that \begin{equation}
\label{em2l}
A(r)=\frac{1}{1-b r^{2}},\tag{4}
\end{equation}
where $b$ is some constant of dimension $L^{-2}$. As result of the equations (1),(2) and (3), the scalar Ricci curvature, pressure and energy density are constant too:
\begin{equation}
\varepsilon=3~b,~~~p=-b,~~~R=-6~b.\tag{5}
\end{equation}
Your metric describes homogeneous, static and isotropic solution derived by Einstein in 1917 called later Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .
