= Solution
At second order in the <interaction term>, attach the two external <fermion> fields to different vertices and contract the remaining two <fermion> fields with each other and the two vector fields with each other. The factor $1/2!$ in the expansion is cancelled by the two ways to assign the external ends. There is no closed <fermion loop>, and hence no extra closed-loop <fermionic sign>.
After removing the two external <Dirac propagators>, the internal <momentum> assignment $l$ for the <photon> and $p+l$ for the <fermion> gives the one-loop insertion
$$
\boxed{\mathcal I(p)=(-ie\mu^{\epsilon/2})^2\int\frac{d^dl}{(2\pi)^d}
\gamma^\nu\widetilde S(p+l)\gamma^\mu\widetilde\Delta_{\mu\nu}(l)
=-e^2\mu^\epsilon\int\frac{d^dl}{(2\pi)^d}
\gamma^\nu\widetilde S(p+l)\gamma^\mu\widetilde\Delta_{\mu\nu}(l).}
$$
Thus the insertion is proportional to exactly the propagator product specified in the source. Define $\mathcal I=-i\Sigma$. With Fourier convention $\psi(x)\sim e^{ip\cdot x}$, the reduced free inverse <kernel> is $K_0(p)=-\not p-m$, since the printed <Dirac propagator> equals $iK_0^{-1}$. Repeated insertions give
$$
S=i(K_0-\Sigma)^{-1},\qquad
\Gamma^{(2)}=K_0-\Sigma+K_{\rm ct}.
$$
The factors of $i$ and the sign of the named <self-energy> are convention dependent; the cancelling <counterterm> is fixed once this insertion convention is specified. The internal <photon> joins two vertices on one directed <fermion> line, as in the first panel of the figure in 1b.
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