= Solution
Expand the quantized field using the normalized mode
$$
u_k(\eta)=A_k\frac{1+ik\eta}{\eta^3}e^{-ik\eta}.
$$
The <canonical commutation relation> fixes the Wronskian $M_{\rm Pl}^2z^2(u_ku_k^{*\prime}-u_k^*u_k')=i$. Writing $x=k\eta$, the primed correlators are
$$
\boxed{\langle\mathcal R\mathcal R\rangle'
=\frac{|A_k|^2}{\eta^6}(1+x^2)},
$$
$$
\boxed{\langle\mathcal R\dot{\mathcal R}\rangle'
=\frac{H|A_k|^2}{\eta^6}
(3+2x^2-ix^3)},
$$
$$
\boxed{\langle\dot{\mathcal R}\dot{\mathcal R}\rangle'
=\frac{H^2|A_k|^2}{\eta^6}
(9+3x^2+x^4)}.
$$
Each unprimed expression includes $(2\pi)^3\delta^{(3)}(\mathbf k+\mathbf k')$. The imaginary part of the mixed correlator gives
$$
|\langle[\mathcal R,\dot{\mathcal R}]\rangle'|
=\frac{2H|A_k|^2}{|\eta|^6}|x|^3.
$$
Consequently the <classicality parameter of a cosmological perturbation> is
$$
\boxed{\mathcal C(k,\eta)=
\frac{2|k\eta|^3}
{\sqrt{(1+k^2\eta^2)(9+3k^2\eta^2+k^4\eta^4)}}}
\simeq\frac23|k\eta|^3.
$$
It tends to zero on superhorizon scales, so the perturbation becomes effectively classical even while its amplitude grows as $a^3$. In ordinary slow-roll inflation, $\mathcal R$ instead freezes to a constant; its analogous normalized commutator also vanishes, but behaves as $|k\eta|$ for the leading de Sitter mode rather than $|k\eta|^3$.
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