= Solution
Define the <Mandelstam variables> for the scattering labels by $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2=q^2$ and $u=(p_1-p_4)^2$. Dropping the <electron> and <muon> <masses> gives
$$
p_1\cdot p_2=p_3\cdot p_4=\frac{s}{2},\qquad
p_1\cdot p_4=p_3\cdot p_2=-\frac{u}{2}.
$$
These identities follow by expanding the corresponding squared <four-momenta>, using $p_i^2=0$ and <four-momentum conservation>. Substitution into the supplied spin-averaged <matrix element> yields
$$
\overline{|\mathcal M|^2}
=\frac{8e^4}{t^2}\left(\frac{s^2}{4}+\frac{u^2}{4}\right)
=\boxed{2e^4\frac{s^2+u^2}{t^2}}.
$$
The $1/4$ initial <spin> average is already included in the given expression; no further averaging factor belongs here.
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