Answer
$2.362\times10^{-10}\;\rm m$
Work Step by Step
To find the length of the bond of $\rm NaCl$, we need to find the distance between each atom of them and the center of the mass of the molecule.
And to do so, we need to find the energies of the three successive wavelengths for rotational transitions.
$$\Delta E_1=\dfrac{hc}{\lambda_1}=\dfrac{6.626\times10^{-34}\cdot 3\times10^8}{23.1\times10^{-3}}$$
$$\Delta E_1=\color{green}{ \bf 8.605\times 10^{-24}}\;\rm J $$
$$\Delta E_2=\dfrac{hc}{\lambda_2}=\dfrac{6.626\times10^{-34}\cdot 3\times10^8}{11.6\times10^{-3}}$$
$$\Delta E_2=\color{green}{ \bf 1.714\times 10^{-23}}\;\rm J$$
$$\Delta E_3=\dfrac{hc}{\lambda_3}=\dfrac{6.626\times10^{-34}\cdot 3\times10^8}{7.71\times10^{-3}}$$
$$\Delta E_3=\color{green}{ \bf 2.578\times 10^{-23}}\;\rm J$$
And since
$$\dfrac{\Delta E_2}{\Delta E_1}=\dfrac{1.714\times 10^{-23}}{8.605\times 10^{-24}}=1.99\approx \color{red}{\bf 2}$$
$$\dfrac{\Delta E_3}{\Delta E_1}=\dfrac{2.578\times 10^{-23}}{8.605\times 10^{-24}}=2.99\approx \color{red}{\bf 3}$$
Thus, the three transitions are representing the rotational energy levels of
- From $l=1$ to $l=0$
- From $l=2$ to $l=1$
- From $l=3$ to $l=2$
Hence, $$\Delta E_1=\dfrac{\hslash^2}{I}$$
So, from the figure below, we can see the center of the mass of the $\rm NaCl$ molecule is close to $\rm Cl$ since it has a greater atomic mass.
Thus, the bond length is given by
$$R=r_1+r_2\tag 1$$
and hence,
$$m_{\rm Na}r_2+m_{\rm Cl}r_2=m_{\rm Na}(0)+m_{\rm cl}(R)$$
$$r_2=\dfrac{m_{\rm Na}(0)+m_{\rm cl}(R)}{m_{\rm Na} +m_{\rm Cl} }$$
Pugging the known;
$$r_2=\dfrac{ 35.5R }{ 23 +35.5}$$
$$r_2=0.607R\tag 2$$
Now we need to find $I$;
$$I=m_{\rm Cl}r_1^2+m_{\rm Na}r_2^2$$
noting, from (1), that $r_1=R-r_2$, and, from (2), that $r_2=0.607R$. Hence, $r_1=R-0.607R=0.393R$
$$I=m_{\rm Cl}(0.607R)^2+m_{\rm Na}(0.607R)^2$$
Plugging the known;
$$I=\left[35.5(0.3937R)^2+23(0.607R)^2\right]\times 1.66\times 10^{-27}$$
$$I=2.317\times10^{-26}R^2\tag 3 $$
Now we know that
$$\Delta E_1=\dfrac{\hslash^2}{I}$$
Plugging from (3), from $\Delta E_1$ above, and the given; then solving for $R$;
$$8.605\times 10^{-24}=\dfrac{(6.626\times 10^{-34})^2}{(2\pi)^22.317\times10^{-26}R^2}$$
Thus,
$$R=\color{red}{\bf 2.362\times10^{-10}}\;\rm m$$
