= Solution
With outgoing physical photon <polarization vectors>, the <Feynman rule> gives
$$
\mathcal A_{\lambda_1\lambda_2}=i g_a\epsilon_{\mu\nu\rho\sigma}q_1^\rho q_2^\sigma
\epsilon_1^{\mu *}\epsilon_2^{\nu *}.
$$
The overall phase does not affect the <decay width>. Replacing either polarization vector by its own <four-momentum> gives zero by antisymmetry. Thus the <Ward identity> permits replacement of each physical <photon polarization sum> by $-\eta^{\mu\nu}$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-\eta_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.
The contraction of the two <Levi-Civita symbols> gives
$$
\sum_{\lambda_1,\lambda_2}|\mathcal A|^2
=2g_a^2\left[(q_1\cdot q_2)^2-q_1^2q_2^2\right]
=\frac{g_a^2m_a^4}{2},
$$
since $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero <axion>. The result is independent of direction in its rest frame. Including the identical-particle factor explained in <6/c/solution> therefore gives
$$
\boxed{\Gamma(a\to\gamma\gamma)=\frac1{2!}\frac{4\pi}{64\pi^2m_a}\frac{g_a^2m_a^4}{2}
=\frac{g_a^2m_a^3}{64\pi}.}
$$
Omitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.<Review of Particle Physics by Particle Data Group (2022)>[Axions and Other Similar Particles, page 4, equation (90.8): two-photon decay width for the coupling multiplying a F dual-F divided by four]{ref}
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