= Solution
In the displayed connected <Feynman diagram>, each sextic vertex carries two slow external legs and the four remaining legs at each vertex are paired across the vertices. The second <cumulant expansion> has a factor $-1/2$, the slow legs can be selected in $\binom62^2$ ways, and the four cross-contractions can be paired in $4!$ ways. The coefficient is therefore
$$
-\frac12\binom62^2 4!=-2700.
$$
Writing $\mathcal S_\zeta=\{q:\Lambda/\zeta<|q|<\Lambda\}$, the local zero-external-momentum contribution is
$$
\boxed{
\Delta g(\zeta)=-2700\,\zeta^{4-d}\lambda_0^2
\int_{\mathcal S_\zeta}\frac{d^dq_1d^dq_2d^dq_3}{(2\pi)^{3d}}
G_0(q_1)G_0(q_2)G_0(q_3)G_0(q_1+q_2+q_3)},
$$
with the integral restricted further so that $q_1+q_2+q_3\in\mathcal S_\zeta$. The three independent internal momenta agree with the diagram's <loop order> $L=4-2+1=3$.
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