= Solution
Join three trigonal BO$_3$ units corner to corner. The resulting \b[alternating six-membered B–O ring has one terminal <oxygen> on each boron]:
$$
\boxed{\left[\mathrm{{-}B(O^-){-}O{-}B(O^-){-}O{-}B(O^-){-}O{-}}\right]^{3-}_{\rm cyclic}.}
$$
The superscript records the total charge; the three explicit O$^-$ labels each carry one share of it. Atom counting gives three B, three bridging O and three terminal O. Boron is three-coordinate and approximately trigonal planar. The original structure diagram shows the ring connectivity and terminal atoms.
A planar arrangement permits the perpendicular empty boron p orbitals to accept <pi donation> from suitable <oxygen> <lone pairs>. The terminal <oxygens> can donate effectively toward one boron, giving their B–O linkages more multiple-bond character and explaining why they are shorter than bridging B–O linkages. A bridging <oxygen> shares its interactions between two boron centres. \b[An approximately planar ring is therefore expected] from the trigonal units and favourable orbital alignment. This reasoning does not require describing the entire <anion> as benzene-like with an assumed six-electron <aromatic ring>; the <oxygen> <lone pairs> and terminal donations must actually be counted before making an <aromaticity> claim.
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