= Solution
Work through terms quadratic in the fermions, with Grassmann-odd <Majorana spinors> $\psi_\mu$ and $\epsilon$. At this order the <spin connection> can be taken torsion-free: eliminating its gravitino-induced torsion changes the action only at four-fermion order. The Christoffel term needed to turn the printed $D_\nu\psi_\sigma$ into a full vector-spinor covariant derivative drops out after contraction with $\gamma^{\mu\nu\sigma}$, since the connection is symmetric in $\nu,\sigma$.
First establish the <Einstein tensor contraction of spin curvature>. Expanding the Clifford product and using the <first Bianchi identity>, its five-gamma and antisymmetric three-gamma curvature contributions vanish, leaving
$$
\gamma^{\mu\nu\rho}R_{\nu\rho ab}\gamma^{ab}
=4R^\mu{}_\lambda\gamma^\lambda-2R\gamma^\mu.
$$
Since $[\nabla_\nu,\nabla_\rho]=R_{\nu\rho ab}\gamma^{ab}/4$, it follows that
$$
\boxed{\gamma^{\mu\nu\rho}\nabla_\nu\nabla_\rho\epsilon
=\frac12G^\mu{}_\lambda\gamma^\lambda\epsilon.}
$$
The factor $1/2$ arises from antisymmetrizing the two derivatives, not from a choice of Einstein-Hilbert normalization.
For the <Einstein-Hilbert action>, the metric variation is
$$
\delta g_{\mu\nu}=\frac\kappa2\bigl(\bar\psi_\mu\gamma_\nu\epsilon+\bar\psi_\nu\gamma_\mu\epsilon\bigr).
$$
Variation with respect to the covariant metric, including the determinant variation and discarding the usual boundary divergence, gives
$$
\delta S_{\rm EH}=-\frac1{2\kappa^2}\int |e|G^{\mu\nu}\delta g_{\mu\nu}\,d^4x
=-\frac1{2\kappa}\int |e|\bar\psi_\mu G^\mu{}_\lambda\gamma^\lambda\epsilon\,d^4x.
$$
No gravitational field equation has been assumed here.
For the <Rarita-Schwinger field> action, the determinant, gamma-matrix and connection variations each multiply a term already quadratic in $\psi$. Since $\delta e$ is bilinear in $\psi,\epsilon$, these contributions have four fermionic factors and are outside the requested order. Retain only the variations of the two explicit gravitino factors. The <four-dimensional Majorana bilinear interchange> gives $\bar\chi\gamma^{\mu\nu\rho}\eta=\bar\eta\gamma^{\mu\nu\rho}\chi$ for odd spinors. After integration by parts, the first gravitino variation is
$$
\frac12\int |e|\overline{\delta\psi_\mu}\gamma^{\mu\nu\rho}\nabla_\nu\psi_\rho
=-\frac12\int |e|\bar\psi_\rho\gamma^{\mu\nu\rho}\nabla_\nu\delta\psi_\mu
=\frac12\int |e|\bar\psi_\mu\gamma^{\mu\nu\rho}\nabla_\nu\delta\psi_\rho,
$$
where the last step swaps $\mu,\rho$ and reverses the three-gamma sign. It is exactly equal to the variation of the second gravitino factor. Therefore
$$
\delta S_{\rm RS}=\int |e|\bar\psi_\mu\gamma^{\mu\nu\rho}\nabla_\nu\delta\psi_\rho
=\frac1\kappa\int |e|\bar\psi_\mu\gamma^{\mu\nu\rho}\nabla_\nu\nabla_\rho\epsilon
=\frac1{2\kappa}\int |e|\bar\psi_\mu G^\mu{}_\lambda\gamma^\lambda\epsilon.
$$
The curvature contribution cancels $\delta S_{\rm EH}$ pointwise. Thus
$$
\boxed{\delta(S_{\rm EH}+S_{\rm RS})=0}
$$
up to boundary terms and the omitted four-fermion-order variations. This directly proves the stated local <supersymmetry> invariance with its printed coefficients. At higher order, the torsion and quartic-fermion completion of <minimal four-dimensional supergravity> is needed; it cannot be inferred by dropping the terms indefinitely.
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