= Solution
An <equilibrium bond length> requires $B_e$, not merely the ground-vibrational-state constant. Extrapolating the linear <vibration-rotation interaction> law gives
$$
B_e=B_0+\frac{\alpha_e}{2}=10.9995+0.0440
=\boxed{11.0435\ \mathrm{GHz}}.
$$
Using integer isotope masses, the <reduced mass> is
$$
\mu=\frac{64\times19}{64+19}\,u=14.65060\,u
=2.43279\times10^{-26}\ \mathrm{kg}.
$$
For two nuclei $I_e=\mu r_e^2$, so
$$
\boxed{r_e=\sqrt{\frac{h}{8\pi^2\mu B_e}}
=1.76737\times10^{-10}\ \mathrm m\simeq176.7\ \mathrm{pm}.}
$$
Using $B_0$ instead gives $r_0\simeq177.1$ pm, an effective vibrational separation. The small difference is precisely why the weaker vibrational progressions matter when the target is equilibrium geometry. Higher terms in $B_v$ and the rigid-rotor approximation limit this estimate's precision.
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