= Solution
Regulate the zero mode by a soft momentum $\mathbf q$ and write the free modes as
$$
\phi_{\mathbf q}=f_q a_{\mathbf q}+f_q^*a_{-\mathbf q}^\dagger,
\qquad
\Pi_{\mathbf q}=g_q a_{\mathbf q}+g_q^*a_{-\mathbf q}^\dagger.
$$
The <Wronskian normalization> $f_qg_q^*-f_q^*g_q=i$ follows from the canonical commutator, while the <power spectrum> is $P(q)=|f_q|^2$. Inserting the creation part of $Q$ on the ket and the annihilation part on the bra, their difference is precisely the Wronskian. For ${\cal O}=\phi(\mathbf k)\phi(\mathbf k')$ this gives, after taking $q\to0$ and setting the irrelevant normalization $c=1$,
$$
\boxed{i\langle[Q,{\cal O}]\rangle
=\frac1{P(0)}
\langle\phi(\mathbf k)\phi(\mathbf k')\phi(\mathbf0)\rangle}.
$$
Equivalently, before removing the regulator, the right-hand side is the soft three-point function divided by $P(q)$.
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