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}
with $b$ some constant of dimension $L^{-2}$. Es result the Ricci scalar $R$  is constant, too.