= Solution
Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt s$. First fix a final <color charge>; its multiplicity will be restored for the inclusive rate. The photon-exchange <scattering amplitude> is, up to an overall phase,
$$
\mathcal M=\frac{e^2Q_q}{s}[\bar v(p_2)\gamma^\mu u(p_1)][\bar u(k_1)\gamma_\mu v(k_2)].
$$
Averaging over the two spins of each incident particle and summing final spins gives
$$
\overline{|\mathcal M|^2}=\frac{e^4Q_q^2}{4s^2}\operatorname{tr}(\not p_2\gamma^\mu\not p_1\gamma^\nu)\operatorname{tr}(\not k_1\gamma_\mu\not k_2\gamma_\nu).
$$
By the <gamma matrix trace identities>, the first trace is $4(p_2^\mu p_1^\nu+p_2^\nu p_1^\mu-g^{\mu\nu}p_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, giving
$$
\overline{|\mathcal M|^2}=\frac{8e^4Q_q^2}{s^2}\bigl[(p_1\cdot k_1)(p_2\cdot k_2)+(p_1\cdot k_2)(p_2\cdot k_1)\bigr].
$$
In the centre-of-mass frame all four energies are $\sqrt s/2$. The first pair of scalar products equals $s(1-\cos\theta)/4$ each, and the second pair equals $s(1+\cos\theta)/4$ each. Hence $\overline{|\mathcal M|^2}=e^4Q_q^2(1+\cos^2\theta)$.
For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal F=4\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal F$, not $1/(4\mathcal F)$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}=s/2$. Use this <massless two-body scattering flux normalization>, keeping the spin average separate.
To obtain the <two-body Lorentz-invariant phase space> directly, integrate the momentum-conserving spatial delta function, leaving $\boldsymbol k_2=-\boldsymbol k_1$. The remaining radial integral is $\int dk\,\delta(\sqrt s-2k)=1/2$, so $d\Phi_2/d\Omega=1/(32\pi^2)$. The <relativistic scattering cross-section> is consequently
$$
\boxed{\frac{d\sigma_{\rm one\ color}}{d\Omega}=\frac{\overline{|\mathcal M|^2}}{64\pi^2s}=\frac{\alpha^2Q_q^2}{4s}(1+\cos^2\theta).}
$$
Since $\int d\Omega(1+\cos^2\theta)=2\pi\int_{-1}^1(1+x^2)\,dx=16\pi/3$, the integrated <electron-positron annihilation into a quark pair> result, for this fixed color, is
$$
\boxed{\sigma_{\rm one\ color}=\frac{4\pi\alpha^2Q_q^2}{3s}.}
$$
The first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible <quark> flavor contributes independently. Away from thresholds and narrow resonances, <quark-hadron duality> relates the inclusive parton rate to the observed <inclusive electron-positron annihilation into hadrons>, giving
$$
\boxed{\sigma_{\rm had}^{(\gamma)}\simeq\frac{4\pi\alpha^2}{3s}\,3\sum_{f\ {
m accessible}}Q_f^2.}
$$
This formula concerns the photon channel, assumes $\sqrt s$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include <Z boson> exchange.
Beyond leading order, an exactly exclusive pair of colored <quarks> is not an observable asymptotic state because of <confinement> and <hadronization>. Moreover its virtual corrections have <infrared divergences>; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar qg$, with their degenerate soft and collinear limits. Real and virtual <infrared divergences> then cancel for an <infrared-safe observable>. A specified jet observable can also be meaningful if it defines an <infrared-safe observable>; a bare two-quark final state is not.
A convenient inclusive calculation uses the <optical theorem>. For the electromagnetic hadronic current $j_\mu=\sum_fQ_f\bar q_f\gamma_\mu q_f$, define
$$
i\int d^4x\,e^{iq\cdot x}\langle0|Tj_\mu(x)j_\nu(0)|0\rangle=(q_\mu q_\nu-q^2g_{\mu\nu})\Pi(q^2).
$$
The <hadronic electromagnetic-current spectral density> is obtained from its discontinuity. In this convention the <hadronic R ratio> is $R(s)=\sigma_{\rm had}/\sigma_{\mu\mu}=12\pi\operatorname{Im}\Pi(s+i0)$, where $\sigma_{\mu\mu}=4\pi\alpha^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative <Quantum chromodynamics> applied to this inclusive quantity implements the required sum of real and virtual contributions.
Finally, write $a=\alpha_s(\mu^2)$, $C=1.99-0.11n_f$, $L=\ln(s/\mu^2)$, and $\beta_0=11-2n_f/3$. The given truncated correction is
$$
K=1+\frac a\pi+\frac{a^2}{\pi^2}\left(C-\frac{\beta_0}{4}L\right).
$$
Differentiate with respect to $t=\ln\mu^2$, using the stated one-loop <beta function (physics)> $da/dt=-\beta_0a^2/(4\pi)$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression gives
$$
\frac{dK}{dt}=-\frac{\beta_0a^3}{2\pi^3}\left(C-\frac{\beta_0}{4}L\right).
$$
Thus the <principle of minimal sensitivity>, applied to this expression with nonzero coupling and $\beta_0\ne0$, selects the <one-loop stationary scale for the hadronic annihilation correction>
$$
\boxed{\mu_*^2=s\exp\left[-\frac{4(1.99-0.11n_f)}{11-2n_f/3}\right].}
$$
This is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every <renormalization scale>; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected <running coupling> remains perturbative. If $a=0$ or $\beta_0=0$, the stated derivative imposes no unique scale.
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