= Solution
For a <connected Feynman diagram> with cubic vertices with four external legs and one loop, $3V=2I+4$ and $1=I-V+1$ give $V=I=4$. In a <one-particle-irreducible Feynman diagram>, every internal edge lies on the single cycle. Each of the four cubic vertices therefore has two internal and one external leg, giving a <box Feynman diagram>. With labelled external momenta there are three inequivalent cyclic orderings, modulo rotation and reversal, represented by $1234$, $1243$ and $1324$:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-304-boxes.png]
{title=The three labelled cubic-scalar box diagrams}
{description=The three inequivalent external-leg orderings of a <box Feynman diagram> in a <six-dimensional cubic scalar field theory>. Each square contains four internal propagators and four cubic vertices.}
Take all momenta incoming with $\sum_i p_i=0$. The <Feynman rules> for the <Euclidean action> give a propagator $(q^2+m^2)^{-1}$ and a cubic insertion $-g$. For order $1234$, the amputated connected insertion is
$$
\mathcal B_{1234}=g^4\int\frac{d^6q}{(2\pi)^6}\frac{\chi(q)\chi(q+p_1)\chi(q+p_1+p_2)\chi(q-p_4)}{(q^2+m^2)((q+p_1)^2+m^2)((q+p_1+p_2)^2+m^2)((q-p_4)^2+m^2)},
$$
where $\chi(q)=\mathbf1_{\{\Lambda<|q|<\Lambda_0\}}$ is the high-mode projector for a strict <Wilsonian effective action>. Each labelled box has <Feynman-diagram symmetry factor> one. A common fixed-loop-cutoff convention instead restricts only the chosen $q$ integration to the shell; both prescriptions give the same zero-external-momentum value. The effective-action vertex has the opposite sign to this connected insertion. Thus $\boxed{\text{three labelled boxes, with one cyclic topology}.}$
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