 Solution

The two one-loop Feynman diagrams are fermion loops with three photon interaction vertices. Fix vertex 1; the two cyclic orders are $(1,2,3)$ and $(1,3,2)$. A cyclic rotation does not produce another Feynman diagram. Keeping all three external labels in the same positions, these two orders can be represented by the same triangle with opposite fermion flow, as in the first two panels.

The photons are labelled $(k_1,\mu )$, $(k_2,\nu )$, $(k_3,\rho )$, with all four-momenta incoming and $k_1+k_2+k_3=0$. The amputation of external propagators removes the three external photon propagators, leaving their Lorentz indices open. There are no other connected one-loop Feynman diagrams with only three external photons in QED.