= Solution
Express the <rotational constant> as a <spectroscopic wavenumber> $\widetilde B=h/(8\pi^2cI)$. The rotational absorption <selection rule> $\Delta J=+1$ gives
$$
\widetilde\nu_{J\to J+1}=2\widetilde B(J+1).
$$
Successive lines therefore have spacing $2\widetilde B$, regardless of the starting $J$. Both observed spacings are $20.88\ \mathrm{cm^{-1}}$, giving \b[$\widetilde B=10.44\ \mathrm{cm^{-1}}$]. The three lines are consequently the $5\to6$, $6\to7$ and $7\to8$ transitions within the stated precision; the first observed line is not the $0\to1$ line.
Using the given isotopic masses, the <reduced mass> is
$$
\mu=\frac{1.0078(34.969)}{1.0078+34.969}\,m_u
=1.62661\times10^{-27}\ \mathrm{kg}.
$$
With $\widetilde B=1044\ \mathrm{m^{-1}}$ and $c$ in $\mathrm{m\,s^{-1}}$, the <bond length> follows from
$$
\boxed{r=\sqrt{\frac{h}{8\pi^2c\mu\widetilde B}}
=1.28390\times10^{-10}\ \mathrm m\simeq1.284\ \text{Å}.}
$$
The small common $0.01\ \mathrm{cm^{-1}}$ offset relative to integer multiples of the spacing is immaterial at the precision of this model. A real ground-vibrational-state <rotational constant> determines an effective vibrationally averaged <bond length>; it need not equal the exact equilibrium separation.
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