= Solution
Let $\pi:G\to G/[G,G]$ be the <abelianization> map. Its image $\pi(A)$ is again a $K$-<approximate group>. Apply the large-progression form of the <Freiman-Green-Ruzsa theorem> to $\pi(A)$. It gives a finite subgroup $H$, elements $x_1,\ldots,x_r$, and lengths $L_1,\ldots,L_r$, with
$$
r\leq K^{O(1)},\qquad
HP(x_1,\ldots,x_r;L_1,\ldots,L_r)\subseteq\pi(A^4),
$$
and
$$
|HP|\geq\exp(-K^{O(1)})|\pi(A)|.
$$
The <large lifted product from a coset progression> applied to this progression gives
$$
\left|
\bigl(A^{16}\cap\pi^{-1}(H)\bigr)
\prod_{i=1}^r\bigl(A^{22}\cap\pi^{-1}(\langle x_i\rangle)\bigr)
\right|
\geq \exp(-K^{O(1)})|A|.
$$
Briefly, choose a section of $\pi$ on $\pi(A^6)$. Multiplication by that section is multiplicative up to $A^{12}\cap[G,G]$; lifting successively the subgroup part and each progression direction therefore places every element of $A^6\cap\pi^{-1}(HP)$ in the displayed product. The <fiber-counting lemma for a quotient map> gives $|A^6\cap\pi^{-1}(HP)|\geq\exp(-K^{O(1)})|A|$, which proves the estimate.
Set
$$
A_0=A^{16}\cap\pi^{-1}(H),\qquad
A_i=A^{22}\cap\pi^{-1}(\langle x_i\rangle)\quad(1\leq i\leq r).
$$
The <intersection of an approximate group power with a subgroup> shows that each $A_i$ is a $K^{O(1)}$-approximate group contained in $A^{O(1)}$. The preimage of a <cyclic subgroup> of $G/[G,G]$ has step less than $s$. The same is true of the preimage of the finite subgroup $H$ because $G$ is a <torsion-free group>. Consequently each $\langle A_i\rangle$ has step less than $s$, and the displayed estimate is the required conclusion.
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