= Solution
A <phi-fourth theory> interaction has <mass dimension> $4-D$, so its bare coefficient must contain a <renormalization scale> factor $\mu^{4-D}=(\mu^2)^{2-D/2}$ when expressed through a dimensionless <renormalized coupling>. The tree-level four-leg <interaction vertex> is linear in that coupling. At the next order, two quartic vertices form the three one-loop <bubble diagram> graphs shown in <2/c>, with external pairings $12|34$, $13|24$ and $14|23$.
For each pairing there are two identical internal lines and <Feynman-diagram symmetry factor> 2. The loop integral has the large-<momentum> behaviour
$$
\int\frac{d^Dq}{(2\pi)^D}\frac1{q^2(q+P)^2}
\sim\int^\infty dq\,q^{D-5}.
$$
It is logarithmically divergent in four dimensions. <Dimensional regularization> represents that local divergence by a simple pole proportional to $1/(D-4)$. Its residue is independent of external <momenta>, so a local quartic <counterterm> cancels it. The three graphs sum to one constant coefficient; this argument establishes its structure without requiring its value.
The relation consequently has the form
$$
\boxed{\lambda_0=\mu^{4-D}\left[g+\frac{Cg^2}{D-4}+O(g^3)\right].}
$$
The sign of $C$ incorporates the chosen pole denominator and vertex convention. In particular, a negative $C$ here corresponds to a positive coefficient of $1/(4-D)$ in the bare coupling. Finite pieces depend on the <renormalization scheme>; a pure-pole relation is natural in <minimal subtraction scheme>. One-loop two-point <tadpole diagrams> contribute <mass> <renormalization>, but not the momentum-dependent <field renormalization> that could change this coupling relation at order $g^2$. The four-point object relevant here is the amputated <one-particle-irreducible vertex>; external-leg <self-energy> insertions are dealt with separately.
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