= Solution
In the printed geometry, one B–F bond lies along the in-plane axis and contributes zero. The other two fluorines have perpendicular distance $R\sin60^\circ=\sqrt3R/2$; boron is again on the axis. Therefore
$$
\boxed{I_b=2m_{\rm F}\left(\frac{\sqrt3R}{2}\right)^2
=\frac32m_{\rm F}R^2=\frac{I_c}{2}.}
$$
The other in-plane principal moment is the same by threefold symmetry. Hence <Boron trifluoride> is an <oblate symmetric top> with $I_c>I_b$ and $\widetilde C=\widetilde B/2$.
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