= Solution
Let $\varepsilon=\hbar\omega$ be the incident laser-photon energy and $\gamma=W/(mc^2)$. The maximum in the usual soft-laser geometry occurs for a head-on encounter followed by a forward outgoing photon along the electron beam. Boost into the initial electron rest frame: the incident energy becomes $\varepsilon_*=\gamma(1+\beta)\varepsilon$. Backscattering there gives the <Compton scattering> recoil relation $\varepsilon_{*,s}=\varepsilon_*/[1+2\varepsilon_*/(mc^2)]$. Boosting this outgoing photon forward to the lab gives
$$
\boxed{\varepsilon_{\max}=\frac{\gamma^2(1+\beta)^2\varepsilon}{1+2\gamma(1+\beta)\varepsilon/(mc^2)}.}
$$
The familiar $4\gamma^2\varepsilon$ limit additionally requires $\gamma\gg1$ and $\varepsilon_*\ll mc^2$. The denominator is the recoil correction, not an optional factor at arbitrary incident energy.
For completeness, if the incidence direction is prescribed rather than optimized, let its unit vector be $\mathbf n$ and the outgoing direction $\mathbf n_s$. Squaring the residual electron four-momentum on its mass shell gives the more general kinematic relation
$$
\varepsilon_s=\frac{\varepsilon(W-c\mathbf p\cdot\mathbf n)}{W-c\mathbf p\cdot\mathbf n_s+\varepsilon(1-\mathbf n\cdot\mathbf n_s)}.
$$
Thus an unspecified crossing angle must not silently be identified with head-on incidence; the boxed result is the maximum for the stated head-on soft-photon arrangement.
Numerically $W=2\,\mathrm{GeV}$ gives $\gamma=3913.89$, and a $1\,\mathrm{cm}$ photon has energy $1.23984\times10^{-4}\,\mathrm{eV}$. The maximum scattered energy is $7.5970\,\mathrm{keV}$, giving
$$
\boxed{\lambda_{\rm scattered}=1.6320\times10^{-10}\,\mathrm m=0.16320\,\mathrm{nm}.}
$$
The rest-frame incident photon is only about one electronvolt, so the small recoil correction is consistent with the amplification estimate.
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