= Solution
The amputated connected four-point function through one loop contains the tree quartic vertex, the quartic-counterterm vertex, and three bubble diagrams in the $s$, $t$, and $u$ channels. The unamputated connected function also has one-loop self-energy and two-point-counterterm insertions on each external propagator, which do not determine $\delta g$.
For channel momentum $P$, the bubble has symmetry factor $1/2$ and the <Feynman parameter> identity gives an integral proportional to
$$
\frac{g^2}{2}\mu^\epsilon
\int_0^1dx\int\frac{d^d\ell}{(2\pi)^d}
\frac1{[\ell^2+\Delta(x,P)]^2}.
$$
In <dimensional regularization>, $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, so the pole from one channel is $g^2/(16\pi^2\epsilon)$. Summing the three crossing channels gives $3g^2/(16\pi^2\epsilon)$. The <minimal subtraction scheme> cancels only this pole, hence
$$
\boxed{\delta g=\frac{3g^2}{16\pi^2\epsilon}}.
$$
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