The correct option is A BF3
In BF3, there is a vacant p-orbital in boron and lone pairs are present in fluorine, So due to presenece of donor and acceptor atoms adjacent to each other and to decrease interelectronic repulsion of lone pairs of fluorine, lone pair donation takes place from fluorine to boron. In H2O, NH3 and HF, O, N and F have lone pairs respectively but there is no vacant orbital in hydrogen to accept lone pairs. Thus no back bonding takes place.
Hence, option (a) is the correct answer.