 Solution

At equilibrium, detailed balance requires zero net probability current along each reversible link: $k_f{p}_A^{\mathrm{eq}}=k_b{p}_B^{\mathrm{eq}}$. The equilibrium probabilities satisfy $p_i^{\mathrm{eq}}\propto e^{-G_i/(k_{B}T)}$, where each state free energy includes its internal degeneracy. ThereforeIn 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 $\boxed{k_f/k_b=e^{\Delta G_{AB}/(k_{B}T)}}$. 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 $2k_{B}T$ gives a forward/backward ratio $e^2$, not $e^{-2}$.