= Solution
{c}
Let $\Sigma=m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single <tree-level Feynman diagram> is annihilation through a <virtual photon>. Up to an irrelevant common phase, its <scattering amplitude> is
$$
\mathcal M_\gamma=\frac{e^2Q_eQ_t}{s}\,[\bar v(p_2)\gamma^\mu u(p_1)]\,[\bar u(p_3)\gamma_\mu v(p_4)]\,\delta_{cd}.
$$
There are two initial <spin angular momentum> states for each particle, so the <spin average> is $1/4$. The final <color charge> sum gives $\sum_{c,d}|\delta_{cd}|^2=N_c$. There is no additional <photon polarization> average for a <virtual photon> and no separate polarization label beyond the <Dirac spinor> <spin sums>.
The <spin sums> and the <gamma-matrix trace> identities give
$$
\begin{aligned}
L^{\mu\nu}&=\operatorname{tr}[(\not p_2-m_e)\gamma^\mu(\not p_1+m_e)\gamma^\nu]\\
&=4[p_2^\mu p_1^\nu+p_2^\nu p_1^\mu-\tfrac{s}{2}g^{\mu\nu}],\\
H_{\mu\nu}&=\operatorname{tr}[(\not p_3+m_t)\gamma_\mu(\not p_4-m_t)\gamma_\nu]\\
&=4[p_{3\mu}p_{4\nu}+p_{3\nu}p_{4\mu}-\tfrac{s}{2}g_{\mu\nu}].
\end{aligned}
$$
Here $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting gives
$$
\frac14L^{\mu\nu}H_{\mu\nu}
=8[(p_1\cdot p_3)(p_2\cdot p_4)+(p_1\cdot p_4)(p_2\cdot p_3)]+4s\Sigma.
$$
Since $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma-t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma-u$, this is the <massive vector-current annihilation kernel>
$$
2[(\Sigma-t)^2+(\Sigma-u)^2+2s\Sigma]
=2[t^2+u^2+4s\Sigma-2\Sigma^2].
$$
Thus the absolutely normalized result is
$$
\boxed{\overline{|\mathcal M_\gamma|^2}
=\frac{2N_ce^4Q_e^2Q_t^2}{s^2}[t^2+u^2+4s\Sigma-2\Sigma^2]
=\frac{8e^4}{3s^2}[t^2+u^2+4s\Sigma-2\Sigma^2].}
$$
After taking out the displayed prefactor, $\boxed{A=1,\ B=4,\ C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the <scattering amplitude>; the vanishing threshold <scattering cross-section> comes from <relativistic two-body phase space>.
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