= Solution
At <thermal equilibrium>, <microscopic reversibility> equates the probabilities of a path and its reversed path when both include their <Boltzmann distribution> initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the <Hamiltonian> is even under <time reversal in classical mechanics>,
$$
\rho_{\rm eq}(z_1)\mathbb P_F
=\rho_{\rm eq}(z_2)\mathbb P_B,
\qquad \rho_{\rm eq}(z)\propto e^{-\beta H(z)},
\qquad \beta=\frac1{k_BT}.
$$
This <detailed balance> condition and the <energy balance for an autonomous Lagrangian> yield
$$
\frac{\mathbb P_F}{\mathbb P_B}=e^{-\beta(H_2-H_1)},
\qquad \frac{2\zeta}{\sigma^2}=\beta,
\qquad \boxed{\sigma^2=2\zeta k_BT}.
$$
This is the <fluctuation-dissipation relation for a Langevin particle>: the strength of <Gaussian white noise> is fixed by the damping and <temperature>, with $k_B$ the <Boltzmann constant>.
For an equilibrium <coarse-grained variable>, the unresolved microscopic states contribute <entropy>; their statistical weight is encoded in the <Helmholtz free energy>, rather than in a single microscopic energy. Relative to the same reference measure, $\rho_{\rm eq}(x)\propto e^{-\beta F(x)}$, so <microscopic reversibility> becomes
$$
\boxed{\frac{\mathbb P_F}{\mathbb P_B}=e^{-\beta(F_2-F_1)}}.
$$
This extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.
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