= Solution
{c}
Use $x^0=ct$ and <metric signature> $(-,+,+,+)$. With the source's <curvature sign convention>, the <Ricci tensor> is
$$
\begin{aligned}
R_{\mu\nu}&=-R^\lambda{}_{\mu\nu\lambda}\\
&=\partial_\nu\Gamma^\lambda_{\mu\lambda}
-\partial_\lambda\Gamma^\lambda_{\mu\nu}
+\Gamma^\sigma_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}
-\Gamma^\sigma_{\mu\nu}\Gamma^\lambda_{\sigma\lambda}.
\end{aligned}
$$
Its time-time component is therefore
$$
R_{00}=\partial_0\Gamma^\lambda_{0\lambda}
-\partial_\lambda\Gamma^\lambda_{00}
+\Gamma^\sigma_{0\lambda}\Gamma^\lambda_{0\sigma}
-\Gamma^\sigma_{00}\Gamma^\lambda_{\sigma\lambda}.
$$
For a slowly moving, pressure-negligible source, the <stress-energy tensor> is approximated by <relativistic dust>: $T^{\mu\nu}=\rho U^\mu U^\nu$, where $\rho$ is <mass density> and $U^\mu$ is the <four-velocity>. In its approximate rest <local inertial frame>, $U^\mu\simeq(c,0,0,0)$ and $U_0\simeq-c$. Thus
$$
T_{00}\simeq\rho c^2,\qquad T_{ij}\ll T_{00},\qquad
T\simeq-\rho c^2,
$$
and the <trace-reversed stress-energy tensor> satisfies
$$
S_{00}=T_{00}-\frac12g_{00}T\simeq\frac12\rho c^2.
\qquad\boxed{R_{00}\simeq\frac12C\rho c^2.}
$$
Independently, introduce the <metric perturbation> $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, with $|h_{\mu\nu}|\ll1$ and constant <Minkowski metric> $\eta_{\mu\nu}$. The <Christoffel symbols> are first order in $h$, so their products are discarded in the <weak-field approximation>. With a comma denoting a <partial derivative>, the linearized <Riemann curvature tensor> is
$$
R_{\lambda\mu\nu\kappa}
=\frac12\left(h_{\lambda\nu,\kappa\mu}
-h_{\mu\nu,\kappa\lambda}
-h_{\lambda\kappa,\nu\mu}
+h_{\mu\kappa,\nu\lambda}\right)+O(h^2).
$$
Taking the <tensor contraction> gives, to this order,
$$
R_{00}=\sum_{i=1}^3R_{i0i0}
=\frac12\sum_{i=1}^3\left(\partial_0^2h_{ii}
-2\partial_0\partial_i h_{0i}+\partial_i^2h_{00}\right).
$$
In the static <Newtonian limit>, all time <partial derivatives> vanish. The constant $\eta_{00}$ also has zero <partial derivatives>, leaving
$$
\boxed{R_{00}\simeq\frac12\nabla^2h_{00}
=\frac12\nabla^2g_{00}.}
$$
Here $\nabla^2=\sum_i\partial_i^2$ is the spatial <Laplace operator> in the approximately Cartesian background coordinates. This sign follows the paper's <curvature sign convention>.
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