= Solution
For the amputated <one-particle-irreducible vertex>, every internal bridge is forbidden. Define the <Feynman integrals>
$$
D(k)=\frac{i}{k^2-m^2+i0},\qquad \int_k=\int\frac{d^dk}{(2\pi)^d},
$$
$$
I_2(P)=\int_kD(k)D(k+P),\qquad
I_3(P,Q,R)=\int_kD(k)D(k+P)D(k-R),\quad P+Q+R=0.
$$
For $E=3,L=1$, the counting identity in part b gives $V_3+2V_4=3$. The only <one-particle-irreducible Feynman diagrams> are a triangle of three cubic vertices and a two-propagator bubble joining one cubic and one quartic vertex. Write $\mathcal V_3$ for the amputated Feynman vertex, including its overall factor of $i$. With all external <momenta> incoming,
$$
\boxed{\mathcal V_{3,\,1\text{-loop}}(p_1,p_2,p_3)
=v_3^3 I_3(p_1,p_2,p_3)
+\frac{v_3v_4}{2}\sum_{i=1}^3 I_2(p_i).}
$$
The triangle has $S=1$ for labelled external legs. The mixed bubble has $S=2$ because its two internal lines may be interchanged. There are three distinct terms, according to which external leg is attached to the cubic vertex; the other two attach to the quartic vertex. No tadpole attached by a bridge is <one-particle irreducible>.
The requested position-space three-point <correlation function> is obtained by restoring the external <propagators> and a <Fourier transform>:
$$
G^{(3)}_{\mathrm{1PI},\,1\text{-loop}}(x,y,z)
=\int_{p_1}\int_{p_2}\int_{p_3}(2\pi)^d\delta^d(p_1+p_2+p_3)
e^{-i(p_1x+p_2y+p_3z)}D(p_1)D(p_2)D(p_3)
\mathcal V_{3,\,1\text{-loop}}.
$$
The tree-level contribution would replace the last factor by $v_3$. This explicitly distinguishes the amputated vertex from the <correlation function> while giving all loop integrations in <momentum> space.
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