= Solution
Define $\mathcal M$ by removing the factor $i(2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4)$ from the connected transition <matrix element>. The single <tree-level Feynman diagram> in <1/b/i> gives
$$
i\mathcal M=(-ig)^2\frac{i}{q^2-m^2+i0}
[\bar u_{s_3}(p_3)u_{s_1}(p_1)]
[\bar u_{s_4}(p_4)u_{s_2}(p_2)],
\qquad q=p_1-p_3.
$$
There is no identical-particle exchange term because the species differ. Each incoming <fermion> has two equally populated <spin angular momentum> states in an unpolarized beam. Their probabilities multiply to $1/4$, giving a <spin average> over the four initial choices. The unobserved final <spin angular momentum> states are distinct outcomes, so their probabilities are added by a <spin sum>, with no final averaging:
$$
\overline{|\mathcal M|^2}=\frac14\sum_{s_1,s_2,s_3,s_4}|\mathcal M|^2.
$$
Taking the complex conjugate of $\bar u_3u_1$ gives $\bar u_1u_3$. The <fermion spin sum> therefore converts each squared bilinear into a <gamma-matrix trace>:
$$
\overline{|\mathcal M|^2}
=\frac{g^4}{4(q^2-m^2)^2}
\operatorname{tr}[(\not p_3+m)(\not p_1+m)]
\operatorname{tr}[(\not p_4+m)(\not p_2+m)].
$$
Using $\operatorname{tr}1=4$, $\operatorname{tr}\gamma^\mu=0$, $\operatorname{tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}$, and the <mass shell> conditions $p_j^2=m^2$, each <gamma-matrix trace> is
$$
4(p_1\cdot p_3+m^2)=4\left(2m^2-\frac{q^2}{2}\right)
=2(4m^2-q^2).
$$
The second equality follows from $q^2=2m^2-2p_1\cdot p_3$; <momentum conservation> gives the identical result for $p_2,p_4$. Hence
$$
\boxed{\overline{|\mathcal M|^2}
=g^4\left(\frac{4m^2-q^2}{m^2-q^2}\right)^2.}
$$
\b[The printed prefactor $g^2$ is incorrect for the stated coupling.] Each of the two <Yukawa interaction> vertices contributes one $g$ to the <scattering amplitude>, so its square is proportional to $g^4$. As a concrete counterexample, take $g=2$, $m>0$, and forward elastic kinematics $p_3=p_1$, $p_4=p_2$. Then $q^2=0$ and the <spin average> is $16g^4=256$, whereas the printed expression gives $16g^2=64$.
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