= Solution
{c}
The exchanged particle is now a <gluon>, and the incoming <up quark> mass replaces $m_e$. Define
$$
\Sigma_u=m_u^2+m_t^2,\qquad F_u=t^2+u^2+4s\Sigma_u-2\Sigma_u^2.
$$
The <Dirac trace> and <spin average> are unchanged. Normalize the fundamental <gauge generators> by $\operatorname{tr}(T^AT^B)=\delta^{AB}/2$. Averaging over the $N_c^2$ initial <color charge> choices and summing the final <color charge> choices gives
$$
C_g=\frac1{N_c^2}\sum_{A,B}\operatorname{tr}(T^AT^B)\operatorname{tr}(T^AT^B)
=\frac{N_c^2-1}{4N_c^2}=\frac29.
$$
Replace the factor $N_ce^4Q_e^2Q_t^2$ from part (ii) by $g_s^4C_g$. Therefore
$$
\boxed{\overline{|\mathcal M_g|^2}=\frac{4g_s^4}{9s^2}F_u.}
$$
Keeping $m_u$ is allowed; if the usual negligible <up quark> mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial <quark> <color charge> states are averaged, whereas final ones are summed.
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