= Solution
\b[$k\simeq5.10\times10^2\ \mathrm{N\,m^{-1}}$ and $\gamma\simeq0.638\ \mathrm{N\,m^{-1}}$.] With $m=m_{\rm u}$ and $\omega_0=2\pi c\widetilde\nu_0$, the dilute-gas stretch gives
$$
k=\frac{m}{2}(2\pi c\,4161.2\ \mathrm{cm}^{-1})^2=510.10\ \mathrm{N\,m^{-1}}.
$$
Use $c$ in $\mathrm{cm\,s^{-1}}$ here. The shifted line is the higher totally symmetric mode; the other molecular stretch remains at $\omega_0$. To first order,
$$
\gamma\simeq m(2\pi c)^2(\widetilde\nu_+^2-\widetilde\nu_0^2)
=\boxed{0.63754\ \mathrm{N\,m^{-1}}}.
$$
The ratio $\gamma/k=1.25\times10^{-3}$ validates the weak-coupling expansion. Inverting the exact eigenvalue instead, with $L=m\omega_+^2$, gives
$$
\gamma=\frac{L(L-2k)}{2(L-k)}=0.63715\ \mathrm{N\,m^{-1}},
$$
a negligible change at the precision of the model.
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