= Solution
Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-<photon> channel is being considered. The spectral-density symbol $\rho_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention is
$$
\begin{aligned}
W_A^{\mu\nu}(q)&=\sum_{X\,\mathrm{hadronic}}\int d\Pi_X\,(2\pi)^4\delta^{(4)}(q-p_X)
\langle0|A^\mu(0)|X\rangle\langle X|A^\nu(0)|0\rangle\\
&=2\pi\rho_h(s)\left(-g^{\mu\nu}+\frac{q^\mu q^\nu}{s}\right),\qquad q^0>0.
\end{aligned}
$$
Here $d\Pi_X$ is the product of final-particle measures $d^3k/[(2\pi)^3 2E]$, including the appropriate sums and symmetry factors. <Lorentz invariance> and <current conservation> give the transverse tensor form. In this convention $\rho_h$ is the <hadronic photon spectral density>, with mass dimension $-2$, rather than a density of a stationary time series.
The electron annihilation amplitude can be written $e\bar v(p_+)\gamma_\mu u(p_-)\langle X|A^\mu|0\rangle$, up to an overall phase. The <leptonic tensor> after the initial spin average is
$$
L_{\mu\nu}=\frac14\operatorname{tr}[\not p_+\gamma_\mu\not p_-\gamma_\nu]
=p_{+\mu}p_{-\nu}+p_{+\nu}p_{-\mu}-g_{\mu\nu}\,p_+\cdot p_-.
$$
It obeys $q^\mu L_{\mu\nu}=0$ and $g^{\mu\nu}L_{\mu\nu}=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt s/2$. Contracting the tensors gives the \b[inclusive <relativistic cross-section>]
$$
\boxed{\sigma_h(s)=\frac{e^2}{2s}L_{\mu\nu}W_A^{\mu\nu}(q)=\pi e^2\rho_h(s)=4\pi^2\alpha\rho_h(s),\qquad\alpha=\frac{e^2}{4\pi}.}
$$
Here $\alpha$ is the <fine-structure constant>. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.
For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^\mu=\sum_fQ_f\bar q_f\gamma^\mu q_f$, and write its inclusive tensor as
$$
W_J^{\mu\nu}=2\pi\rho_J(s)(q^\mu q^\nu-sg^{\mu\nu}).
$$
Equivalently, for $i\int d^4x\,e^{iqx}\langle0|T J_h^\mu(x)J_h^\nu(0)|0\rangle=(q^\mu q^\nu-q^2g^{\mu\nu})\Pi_J(q^2)$, one has $\rho_J=\operatorname{Im}\Pi_J/\pi$. At leading order in the electromagnetic interaction, $\langle X|A^\mu|0\rangle$ is proportional to $e\langle X|J_h^\mu|0\rangle/s$. Thus
$$
\rho_h(s)=\frac{e^2}{s}\rho_J(s),\qquad
\boxed{\sigma_h(s)=\frac{16\pi^3\alpha^2}{s}\rho_J(s).}
$$
If the symbol $\rho_h$ is instead used for this dimensionless <hadronic electromagnetic-current spectral density>, the last formula applies with $\rho_J$ renamed $\rho_h$. The normalization must not be silently switched between these formulas.
At $\sqrt s$ well above the <strong-coupling scale> of <Quantum chromodynamics>, <asymptotic freedom> makes production over distances of order $1/\sqrt s$ perturbative. The electromagnetic current initially creates a <quark>-<antiquark> pair. Subsequent strong interactions produce <hadrons>, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which <quark-hadron duality> motivates a leading <parton model> calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.
For one active massless <quark> flavor with charge $Q_fe$, the tree-level <scattering amplitude> is
$$
\mathcal M_f=\frac{e^2Q_f}{s}[\bar v(p_+)\gamma_\mu u(p_-)][\bar u(k_q)\gamma^\mu v(k_{\bar q})].
$$
The two <gamma-matrix traces>, the initial spin average, and the final <color charge> sum yield
$$
\overline{\sum}|\mathcal M_f|^2=2N_ce^4Q_f^2\frac{t^2+u^2}{s^2}
=N_ce^4Q_f^2(1+\cos^2\theta),
$$
where $t=-s(1-\cos\theta)/2$ and $u=-s(1+\cos\theta)/2$. The massless <two-body Lorentz-invariant phase space> then gives
$$
\frac{d\sigma_f}{d\Omega}=\frac{N_c\alpha^2Q_f^2}{4s}(1+\cos^2\theta),\qquad
\int d\Omega\,(1+\cos^2\theta)=\frac{16\pi}{3}.
$$
Summing the distinct final flavors, rather than interfering amplitudes for them, gives
$$
\boxed{\sigma_h^{(0)}(s)=\frac{4\pi\alpha^2}{3s}N_c\sum_{f\,\mathrm{active}}Q_f^2.}
$$
According to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\geq4m_f^2$.
The \b[<hadronic R ratio>] is $R=N_c\sum Q_f^2$, relative to the massless muon-pair <relativistic cross-section> $4\pi\alpha^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,
$$
R=\frac{N_c}{9}(4N_u+N_d).
$$
For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives
$$
\boxed{\rho_J^{(0)}(s)=\frac{N_c\sum_fQ_f^2}{12\pi^2},\qquad
\rho_h^{(0)}(s)=\frac{\alpha N_c\sum_fQ_f^2}{3\pi s}.}
$$
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