= Solution
For the interaction sign in this paper, the cubic <Feynman rule> is $+i\lambda$, while the scalar <Feynman propagator> is $i/(k^2-m^2+i\epsilon)$. Each exchange graph therefore contributes $-i\lambda^2/(k^2-m^2+i\epsilon)$ to $i\mathcal M$. The three graphs give
$$
\mathcal M=-\lambda^2\left(\frac1{s-m^2+i\epsilon}+\frac1{t-m^2+i\epsilon}+\frac1{u-m^2+i\epsilon}\right).
$$
The equal-mass <Mandelstam variables> satisfy $s+t+u=4m^2$, so $u-m^2=3m^2-s-t$. In the physical elastic region with $m>0$, none of these denominators vanishes, and the tree amplitude is real up to its irrelevant overall convention. Thus
$$
\boxed{|\mathcal M|^2=\lambda^4\left[\frac1{s-m^2}+\frac1{t-m^2}+\frac1{3m^2-s-t}\right]^2.}
$$
The square includes interference between all three channels; summing their individual squared contributions would omit physical cross terms. If a denominator approaches a pole, retain the <Feynman i-epsilon prescription> in the modulus before taking a limit.
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