= Solution
Four external $\Phi$ lines require four cubic <interaction vertices>, since each such <interaction vertex> contains only one heavy field. Their other eight endpoints must form four light internal lines. A <connected Feynman diagram> therefore has $V=I=4$ and $L=1$: the leading contribution is a one-loop <box diagram>, of order $g^4$, with four $\phi$ <scalar propagators>. Charge-two incoming and outgoing heavy lines attach alternately around the box, as shown in panel (d) of <1/b>'s figure. Each incoming heavy line supplies the two outgoing light-charge arrows at its corner, and each outgoing heavy line absorbs them. With external labels fixed there is one such box and its <symmetry factor> is one.
For $|\mathbf p|\ll M\ll m$, the external energies are also small compared with $m$, so expand the loop integrand in all external <four-momenta>. Its leading term is
$$
i\mathcal M_{\Phi\Phi}=g^4\int\frac{d^4k}{(2\pi)^4}\frac1{(k^2-m^2+i\epsilon)^4}
+O\left(\frac{i g^4M^2}{m^6}\right).
$$
A <Wick rotation> and the <Euclidean massive loop integral> give
$$
\int\frac{d^4k}{(2\pi)^4}\frac1{(k^2-m^2+i\epsilon)^4}
=i\int\frac{d^4k_E}{(2\pi)^4}\frac1{(k_E^2+m^2)^4}
=\frac{i}{96\pi^2m^4}.
$$
Hence, more precisely than just its requested scaling,
$$
\boxed{\mathcal M_{\Phi\Phi}=\frac{g^4}{96\pi^2m^4}
\left[1+O\left(\frac{M^2}{m^2},\frac{|\mathbf p|^2}{m^2}\right)\right].}
$$
The <mass dimension> $[g]=1$ makes $g^4/m^4$ dimensionless, as a four-dimensional <scattering amplitude> should be. The leading amplitude has no extra identical-particle multiplicity.
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