= Solution
For $\alpha/\lambda\ll1$, follow the unstable root near the clean-film value $L=\lambda s(1-s)$. Write $\beta=L+\Delta$ in $(\beta-L)(\beta+S)+3LS/4=0$. At first order in $S$, $L\Delta+3LS/4=0$, giving
$$
\beta=\lambda s(1-s)-\frac34\alpha s+O(\alpha^2/\lambda).
$$
This expansion is uniform near the maximizing <wavenumber>. Differentiation gives $s_*=1/2-3\alpha/(8\lambda)+O((\alpha/\lambda)^2)$. Evaluating the growth there yields
$$
\boxed{\beta_{\max}=\frac\lambda4-\frac{3\alpha}{8}+O(\alpha^2/\lambda).}
$$
The change in maximizing <wavenumber> only affects the maximum at second order. The first-order decrease therefore has exactly the requested coefficient, relative to the clean-film rate with its <surface tension> replaced by $\bar\sigma$.
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