= Solution
{c}
Use <Minkowski metric> $g=\operatorname{diag}(1,-1,-1,-1)$ and the convention $S_{fi}=\delta_{fi}+i(2\pi)^4\delta^{(4)}(p_f-p_i)\mathcal M_{fi}$. Put $m=m_t$, $y=\lambda_t/\sqrt2$, $Q=e_t=2/3$ and
$$
D_0=l^2-m^2+i0,\qquad D_1=(l+p_1)^2-m^2+i0,\qquad D_2=(l-p_2)^2-m^2+i0.
$$
The <Feynman i-epsilon prescription> is understood in subsequent denominators. Contracting the <spinor> indices around the <fermion loop> gives the <trace>
$$
T_{\mu\nu}(l)=\operatorname{tr}\!\left[\gamma_\mu(\not l+m)\gamma_\nu(\not l-\not p_2+m)(\not l+\not p_1+m)\right].
$$
Here $\not p=\gamma^\rho p_\rho$ is <Feynman slash notation>. The <closed fermion loop sign> is $-1$, the three <interaction vertices> contribute $(-iy)(-ieQ)^2$, and the three <Dirac propagators> contribute $i^3$. Their product, including the <color charge> trace, is $-N_cye^2Q^2$. Thus one ordering has
$$
i\mathcal M^{(1)}_{\mu\nu}=-N_cye^2Q^2\int\frac{d^4l}{(2\pi)^4}\frac{T_{\mu\nu}(l)}{D_0D_1D_2},\qquad
\boxed{\mathcal M^{(1)}_{\mu\nu}=iN_cye^2Q^2\int\frac{d^4l}{(2\pi)^4}\frac{T_{\mu\nu}(l)}{D_0D_1D_2}.}
$$
Multiplying by the incoming <polarization vectors> and adding the exchanged <Feynman diagram> gives
$$
\boxed{\mathcal M=\epsilon_1^\mu\epsilon_2^\nu\left[\mathcal M^{(1)}_{\mu\nu}(p_1,p_2)+\mathcal M^{(1)}_{\nu\mu}(p_2,p_1)\right].}
$$
The displayed four-dimensional <Feynman integral> is formal until regularized; subsequent manipulations use <dimensional regularization>, which permits the required <loop momentum> shifts.
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