= Solution
Let
$$
G_>(x-y)=\int_{\Lambda/\zeta<|q|<\Lambda}
\frac{d^dq}{(2\pi)^d}\frac{e^{iq\cdot(x-y)}}{q^2+\mu_0^2}.
$$
At order $g_0^2$, the connected six-point topology consists of two quartic vertices joined by one fast line. The second cumulant gives
$$
-8g_0^2\int d^dx\,d^dy\,
\phi_<^3(x)G_>(x-y)\phi_<^3(y).
$$
For a sharp shell and external slow momenta approaching zero, the connecting momentum cannot lie in the fast shell, so this nonlocal tree topology gives no local zero-momentum $\phi_<^6$ coupling.
At order $g_0^3$, the local connected contribution is the triangle of three quartic vertices, with two external slow legs at each vertex and one fast propagator along each edge. The factor is
$$
\frac1{3!}\binom42^3(2^2)=144.
$$
Consequently, at zero external momentum,
$$
\Delta\gamma
=144g_0^3I_3(\zeta),
\qquad
I_3(\zeta)=\int_{\Lambda/\zeta<|q|<\Lambda}
\frac{d^dq}{(2\pi)^d}\frac1{(q^2+\mu_0^2)^3}.
$$
Ignoring contributions involving $\lambda_0,\gamma_0$ and anomalous field rescaling, the running sextic coupling is therefore
$$
\boxed{
\gamma(\zeta)=\zeta^{6-2d}
\left[\gamma_0+144g_0^3I_3(\zeta)\right]+O(g_0^4).}
$$
The diagrams are respectively the one-line two-vertex six-point tree and the three-vertex triangular one-loop six-point diagram; only the latter contributes to the local shell coupling at vanishing external momenta.
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