= Solution
Let $\rho_i=N_i/V$ be a molecular <number density> and define the volume-independent <molecular partition function density> $f_i=q_i/V$. Then $q_i/N_i=f_i/\rho_i$. Also define the reaction reference-energy difference
$$
\Delta\varepsilon_0=\varepsilon^{0,\mathrm C}-\varepsilon^{0,\mathrm A}-\varepsilon^{0,\mathrm B}.
$$
Substitution into the equilibrium chemical-potential relation gives
$$
\begin{aligned}
0&=-k_{\mathrm B}T\ln\frac{f_{\mathrm C}}{\rho_{\mathrm C}}
+k_{\mathrm B}T\ln\frac{f_{\mathrm A}}{\rho_{\mathrm A}}
+k_{\mathrm B}T\ln\frac{f_{\mathrm B}}{\rho_{\mathrm B}}
+\Delta\varepsilon_0,\\
\frac{\rho_{\mathrm C}}{\rho_{\mathrm A}\rho_{\mathrm B}}
&=\frac{f_{\mathrm C}}{f_{\mathrm A}f_{\mathrm B}}
\exp\left(-\frac{\Delta\varepsilon_0}{k_{\mathrm B}T}\right).
\end{aligned}
$$
An <equilibrium constant> expressed using concentration activities $a_i=\rho_i/c^\circ$ is dimensionless:
$$
\boxed{K_c=\frac{a_{\mathrm C}}{a_{\mathrm A}a_{\mathrm B}}
=\frac{\rho_{\mathrm C}c^\circ}{\rho_{\mathrm A}\rho_{\mathrm B}}
=\frac{f_{\mathrm C}c^\circ}{f_{\mathrm A}f_{\mathrm B}}e^{-\Delta\varepsilon_0/(k_{\mathrm B}T)}.}
$$
Here every $f_i$ has units $\mathrm{m^{-3}}$, and $c^\circ$ is the chosen <standard concentration>, initially one molecule per cubic metre. The factor $c^\circ$ is essential: without it the association quotient has units of volume.
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