= Solution
For a connected cubic <Feynman diagram> with $V$ internal <vertices>, $I$ internal lines and three external lines, $3V=2I+3$. Hence $V$ is odd and the <loop order> is $L=I-V+1=(V-1)/2$. Through order $\lambda^3$, there is one <tree-level Feynman diagram> at $V=1$ and seven labelled <one-loop Feynman diagrams> at $V=3$:
* One triangle, with one external line at each <vertex>. Fixing the three external labels leaves no internal automorphism, so its <symmetry factor> is $S=1$.
* Three <bubble diagrams> on external branches of the tree, one for each choice of the distinguished external label $i$. The other two external lines meet the remaining tree <vertex>. Exchanging the two parallel lines gives <symmetry factor> $S=2$ for each labelled graph.
* Three <tadpole diagrams> at the end of a chain. The middle <vertex> carries the distinguished external label $i$, the other two external lines meet one end <vertex>, and the other end bears the loop. Reversing the two half-edges of the loop gives <symmetry factor> $S=2$ for each labelled graph.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2016-cubic-three-point-diagrams.png]
{title=All connected cubic three-point diagrams through third order: one tree and seven labelled one-loop graphs; a diagram of symmetry factor S has coefficient 1/S}
For an explicit check, use the <propagator> notation of <3/b/solution> and define, with $\{i,j,k\}=\{1,2,3\}$,
$$
\begin{aligned}
\mathcal T&=\int d^Dx\,d^Dz\,d^Dw\,
K_1(x)K_2(z)K_3(w)\Delta_{xz}\Delta_{zw}\Delta_{wx},\\
\mathcal B_i&=\int d^Dx\,d^Dz\,d^Dw\,
K_i(z)K_j(w)K_k(w)\Delta_{xz}^2\Delta_{xw},\\
\mathcal A_i&=\int d^Dx\,d^Dz\,d^Dw\,
K_i(z)K_j(w)K_k(w)\Delta_{xz}\Delta_{zw}\,t.
\end{aligned}
$$
There is no ordering of $j,k$ in this notation. The complete connected <correlation function> through this order is
$$
\boxed{G_{3,c}=(-i\lambda)\int d^Dx\,K_1(x)K_2(x)K_3(x)
+(-i\lambda)^3\left[\mathcal T+\frac12\sum_{i=1}^3(\mathcal B_i+\mathcal A_i)\right]+O(\lambda^5).}
$$
For completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct <Wick contraction> counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by $3!(3!)^3=1296$.
\b[Connected does not mean one-particle irreducible.] After <amputation of external propagators>, only the triangle is <one-particle irreducible> at this order; the six other loop graphs belong to the requested connected answer. The <tadpole diagrams> cannot be discarded without an additional subtraction prescription.
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