= Solution
At <equilibrium>, <detailed balance> requires zero net <probability> current along each reversible link: $k_fp_A^{\rm eq}=k_bp_B^{\rm eq}$. The <equilibrium> probabilities satisfy $p_i^{\rm eq}\propto e^{-G_i/(k_BT)}$, where each state <free energy> includes its internal degeneracy. Therefore
$$
\boxed{\frac{k_f}{k_b}=\frac{p_B^{\rm eq}}{p_A^{\rm eq}}
=e^{-(G_B-G_A)/(k_BT)}.}
$$
In the source sketch B is lower, and the labelled vertical drop is naturally $\Delta G_{AB}=G_A-G_B>0$. With that drop convention the ratio is $\boxed{k_f/k_b=e^{\Delta G_{AB}/(k_BT)}}$. If instead $\Delta G_{AB}$ means the conventional reaction change $G_B-G_A$, the exponent has the minus sign. For example a drop of $2k_BT$ gives a forward/backward ratio $e^2$, not $e^{-2}$.
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