= Solution
Factor the momentum dependence into
$$
\mathcal G=(2\pi)^3\delta^3(\sum\mathbf k_i)\frac{S}{k_1^3k_2^3k_3^3}\left(-K+\frac{s_2}{K}+\frac{s_3}{K^2}\right).
$$
Write $s=+1$ for the intended target, obtained with the Hamiltonian sign repair in part (i), and $s=-1$ for the literal stated Hamiltonian. The same sign choice applies to both limits and cancels in their amplitude ratio. The <primordial bispectrum> contribution is $sH^4(\epsilon+1-c_s^2)\mathcal G/(16\epsilon^2c_s^4)$. Hence
$$
\boxed{c_s^2\to1:\quad\langle\zeta^3\rangle_c\simeq s\frac{H^4}{16\epsilon}\mathcal G,\qquad c_s^2\ll1:\quad\langle\zeta^3\rangle_c\simeq s\frac{H^4(1+\epsilon)}{16\epsilon^2c_s^4}\mathcal G.}
$$
At fixed $H,\epsilon$ and momentum triangle, the ratio to the unit-sound-speed amplitude is
$$
\boxed{\frac{|B(c_s)|}{|B(1)|}=\frac{\epsilon+1-c_s^2}{\epsilon c_s^4}\simeq\frac{101}{c_s^4}\quad(\epsilon=0.01,\ c_s^2\ll1).}
$$
Keeping only the leading slow-roll order makes this $100/c_s^4$. To compare the two pieces of this same interaction at a fixed $c_s$, their ratio is instead $(1-c_s^2)/\epsilon\simeq100(1-c_s^2)$. Thus the noncanonical piece overtakes the slow-roll piece once $1-c_s^2\gg0.01$, while both have the common raw $c_s^{-4}$ enhancement from the interaction and its external mode normalization. For instance $c_s^2=0.1$ gives an exact ratio $9100$ to the canonical limit with these fixed parameters.
A raw <primordial bispectrum> amplitude is not the same as normalized <primordial non-Gaussianity>: here $P_\zeta(k)=H^2/(4\epsilon c_sk^3)$, so the ratio $B/P_\zeta^2$ scales as $(\epsilon+1-c_s^2)/c_s^2$ at fixed shape. The small-sound-speed enhancement in normalized non-Gaussianity is therefore $c_s^{-2}$. These are tree-level weak-interaction predictions; approaching arbitrarily small $c_s$ eventually requires checking the perturbative regime.
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