The correct option is A 66.66%
The ground state electronic configuration of B atom is: 1s22s22p1
The excited state electronic configuration of B atom is: 1s22s12p2
Thus, here 2s orbital and two 2p orbitals undergoes hybridisation to give three sp2 hybridised orbitals. Hence, boron in BH3 is sp2 hybridised.
Thus, the percentage of p − character of the hybridised orbital of ‘B′ in BH3 molecule is 23×100=66.66 %