In $B{F}_3$, the $\mathrm{B}-\mathrm{F}$ bond length is $1.30Å$, when $B{F}_3$ is allowed to be treated with $M{e}_{3}N$ , it forms an adduct $M{e}_{3}N\to B{F}_3$. The bond length of $\mathrm{B}-\mathrm{F}$ in the adduct is :

 

Back bonding between fluorinc and boron occurs in ${\mathrm{BF}}_3$ because of the presence of the p-orbital in boron. Back bonding results in the double bond qualities. As ${\mathrm{BF}}_3$ becomes adjuct, the double bond feature disappears since back bonding is no longer present. In turn, this causes the bond ($1.30Å$) to lengthen a little bit.
The right answer is A.