Lewis acid: $BF_3$ Lewis base: $(CH_3)_3N$
Work Step by Step
Since "N" has 5 electrons in its valence shell, and it is doing only 3 ligations , there will be one pair of electrons disposable for donating. In the other side, $B$ has 3 electrons in its valence shell, and it is doing 3 ligations, so it is capable of receiving one pair, to have 8 electrons. The $(CH_3)_3N$ donates one pair of electrons, so it is the Lewis base, and the $BF_3$, who receives, is a Lewis acid.