= Solution
Use ground-state references and the supplied ideal-gas, rigid-rotor and harmonic-oscillator approximations. The two ions have closed electronic shells; take their electronic ground-state degeneracies and the molecular electronic ground-state degeneracy as one, and neglect higher electronic levels. Consistently treated nuclear-spin factors cancel between the species. Since the molecule is heteronuclear, its <rotational symmetry number> is one.
Let $\mu_r=m_{\rm Cs}m_{\rm I}/m_{\rm CsI}$ be the <reduced mass>, neglecting the negligible mass correction from electron transfer. The ratio of <translational partition functions> per volume reduces to
$$
\frac{f_{{\rm trans},{\rm Cs}^+}f_{{\rm trans},{\rm I}^-}}
{f_{{\rm trans},{\rm CsI}}}
=\left(\frac{2\pi\mu_r k_BT}{h^2}\right)^{3/2},
\qquad \frac{\mu_r}{m_u}=\frac{133\times127}{260}=64.9654.
$$
The molecule has additional rotational and vibrational states, whereas the monatomic ions do not. At $3000\,\mathrm K$,
$$
q_{\rm rot}=\frac{3000}{0.0340}=8.82353\times10^4,
\qquad
q_{\rm vib}^0=\frac1{1-e^{-172/3000}}=17.9466,
$$
$$
\left(\frac{2\pi\mu_r k_BT}{h^2}\right)^{3/2}
=1.61698\times10^{34}\,\mathrm{m^{-3}}.
$$
Hence the dimensional equilibrium quotient is
$$
C=\frac{f_{{\rm Cs}^+}f_{{\rm I}^-}}{f_{{\rm CsI}}}
e^{-\Delta\varepsilon_0/(k_BT)}
=\frac{1.61698\times10^{34}}{(8.82353\times10^4)(17.9466)}
(4.87497\times10^{-8})\,\mathrm{m^{-3}},
$$
$$
\boxed{C=4.97797\times10^{20}\,\mathrm{m^{-3}}},\qquad
\boxed{K_c=\frac{4.97797\times10^{20}\,\mathrm{m^{-3}}}{c^\circ}}.
$$
The last form makes the unassigned <standard concentration> explicit. For the common choice of $1\,\mathrm{mol\,dm^{-3}}$, the corresponding microscopic reference is $c^\circ=6.02214\times10^{26}\,\mathrm{m^{-3}}$, giving
$$
\boxed{K_c\simeq8.27\times10^{-7}\quad(c^\circ=1\,\mathrm{mol\,dm^{-3}})}.
$$
Equivalently, the dimensional molar quotient is $8.27\times10^{-7}\,\mathrm{mol\,dm^{-3}}$. A pressure-based gas standard is a different convention: taking $c^\circ=p^\circ/(k_BT)$ with $p^\circ=1\,\mathrm{bar}$ would give $2.06\times10^{-4}$, rather than the same numerical constant.
The well-minimum convention gives the identical answer because
$$
\frac{e^{-(D_e+I-\mathrm{EA})/(k_BT)}}
{e^{-\theta_{\rm vib}/(2T)}q_{\rm vib}^0}
=\frac{e^{-[D_e+I-\mathrm{EA}-k_B\theta_{\rm vib}/2]/(k_BT)}}{q_{\rm vib}^0}.
$$
Constants used here are $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$, $k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}$, $m_u\simeq1.66053907\times10^{-27}\,\mathrm{kg}$ and $N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$.<CODATA recommended values of the fundamental physical constants: 2022 by National Institute of Standards and Technology>[fundamental-constant tables: Planck constant, Boltzmann constant, atomic mass constant and Avogadro constant]{ref}
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