= Solution
Use <metric signature> $(+---)$ and average over the four initial <spin> states, summing over final <spins> and the $N_c=3$ <quark> <color charge> states. The <electrons> carry no <color charge>, so their initial <color charge> average is trivial. A final <color charge> average would not give the inclusive <relativistic cross-section> requested.
<Photon> exchange gives the tree amplitude, up to an irrelevant overall phase,
$$
\mathcal M=\frac{e^2Q_q}{s}\delta_{ab}
[\bar v(p_1)\gamma^\mu u(p_2)]
[\bar u(q_1)\gamma_\mu v(q_2)],\qquad e^2=4\pi\alpha.
$$
Massless <spin sums> give $\sum u\bar u=\not p$ and $\sum v\bar v=\not p$. Therefore
$$
\overline{|\mathcal M|^2}
=\frac{N_c e^4Q_q^2}{4s^2}
\operatorname{tr}(\not p_1\gamma^\mu\not p_2\gamma^\nu)
\operatorname{tr}(\not q_1\gamma_\mu\not q_2\gamma_\nu).
$$
The <color charge> multiplicity is $\sum_{a,b}|\delta_{ab}|^2=N_c$, not $N_c^2$. Each <trace> equals $4(a^\mu b^\nu+a^\nu b^\mu-g^{\mu\nu}a\cdot b)$. Contracting them, the terms proportional to $(p_1\cdot p_2)(q_1\cdot q_2)$ cancel, leaving $32[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]$. Thus
$$
\boxed{\overline{|\mathcal M|^2}
=\frac{384\pi^2\alpha^2Q_q^2}{s^2}
[(p_1\cdot q_2)(p_2\cdot q_1)+(p_2\cdot q_2)(p_1\cdot q_1)]}.
$$
In the <centre-of-momentum frame> every energy is $\sqrt s/2$. Let $\theta$ be the angle between $p_1$ and $q_2$. Then
$$
p_1\cdot q_2=p_2\cdot q_1=\frac s4(1-\cos\theta),\qquad
p_1\cdot q_1=p_2\cdot q_2=\frac s4(1+\cos\theta).
$$
Consequently $\overline{|\mathcal M|^2}=N_c e^4Q_q^2(1+\cos^2\theta)$.
The supplied <two-body Lorentz-invariant phase space> formula has a factor-of-two error. Deriving the normalization explicitly,
$$
d\Phi_2=\frac{d\Omega}{32\pi^2},\qquad
\text{incident flux}=4p_1\cdot p_2=2s,
$$
so the <massless two-body scattering flux normalization> is
$$
\frac{d\sigma}{d\Omega}
=\frac{\overline{|\mathcal M|^2}}{64\pi^2s}
=\frac{\overline{|\mathcal M|^2}}{128\pi^2p_1\cdot p_2}.
$$
For example, integrating the spatial <Dirac delta> in $d\Phi_2$ leaves $[4(2\pi)^2]^{-1}dk\,d\Omega\,\delta(\sqrt s-2k)$; its radial integral supplies the factor $1/2$. The printed denominator $64\pi^2p_1\cdot p_2$ would double the <relativistic cross-section> and cannot reproduce the requested result with the standard <scattering amplitude>.
The correctly normalized differential and total results for <electron-positron annihilation into a quark pair> are
$$
\frac{d\sigma}{d\Omega}=\frac{3\alpha^2Q_q^2}{4s}(1+\cos^2\theta),\qquad
\int d\Omega(1+\cos^2\theta)=\frac{16\pi}{3},
$$
and hence
$$
\boxed{\sigma_q=\frac{4\pi\alpha^2}{s}Q_q^2}.
$$
At energies where <photon> exchange and the <parton> description apply, sum over kinematically accessible <quark> flavors. The produced <quarks> undergo <hadronization>, so the leading inclusive hadronic rate is
$$
\sigma_{\rm had}=\frac{4\pi\alpha^2}{s}\sum_qQ_q^2,\qquad
R(s)=\frac{\sigma_{\rm had}}{\sigma(e^+e^-\to\mu^+\mu^-)}
=3\sum_qQ_q^2.
$$
This is the <hadronic R ratio>. Perturbative <QCD> adds a leading relative correction $\alpha_s(s)/\pi$; thresholds, narrow resonances and, near the <Z boson> scale, electroweak exchange require additional treatment. The calculation is a <photon>-exchange, massless result from a <tree-level Feynman diagram>.
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