The total number of $\mathrm{B}-\mathrm{O}$ sigma bonds in one $[\mathrm{B}{}_4\mathrm{O}{}_5(\mathrm{OH}){}_4]{}^{2-}$ ion (two tetrahedral B atoms and two trigonal B atoms, five bridging oxide O atoms, four terminal OH groups) is

Each tetrahedral B has 4 B-O bonds and each trigonal B has 3, so the sum over boron is 2(4) + 2(3) = 14. Check from oxygen: five bridging O each give 2 bonds (10) and four OH each give 1 (4), total 14.