Another way to get the same answer is to switch everything to frequency units before touching the moment of inertia formula, instead of carrying $B$ in $\mathrm{cm^{-1}}$ all the way through.
The rotational lines of a rigid diatomic sit at $\tilde\nu = 2B(J+1)$ in wavenumber units, and the paper tells us consecutive lines are $41.11\ \mathrm{cm^{-1}}$ apart, so $B = 20.555\ \mathrm{cm^{-1}}$.
Turn this into an ordinary frequency using $\nu = cB$ (with $c$ in $\mathrm{cm\,s^{-1}}$):
\[ \nu = (2.998\times10^{10}\ \mathrm{cm\,s^{-1}})(20.555\ \mathrm{cm^{-1}}) = 6.1638\times10^{11}\ \mathrm{Hz} \]The rotational energy levels are $E_J = \frac{\hbar^2}{2I}J(J+1)$, which turns the same spacing rule into $\nu = h/(4\pi^2 I)$ in ordinary frequency form. Rearranging for $I$:
\[ I = \frac{h}{4\pi^2\nu} = \frac{6.626\times10^{-34}}{4\pi^2\times6.1638\times10^{11}} = 1.3618\times10^{-47}\ \mathrm{kg\,m^2} \]This is the same moment of inertia as before, as it must be, since it is just a different bookkeeping of the same $B$.
Now find the reduced mass in kg using the atomic masses and the amu-to-kg conversion together:
\[ \mu = \frac{(1.008)(18.998)}{1.008+18.998}\times1.661\times10^{-27} = 0.9572\times1.661\times10^{-27} = 1.5899\times10^{-27}\ \mathrm{kg} \]Since $I=\mu r^2$,
\[ r = \sqrt{\frac{I}{\mu}} = \sqrt{\frac{1.3618\times10^{-47}}{1.5899\times10^{-27}}} = 9.25\times10^{-11}\ \mathrm{m} = 0.93\ \mathrm{\mathring{A}} \]The H-F bond length comes out to $0.93\ \mathrm{\mathring{A}}$, the same value reached through the wavenumber route, a good check that both approaches agree.