Ordinarily, we would predict the order to be C > N > O > F.
C is the least electronegative atom.
It has the least tendency to hold on to its electrons, so it is the best electron donor and the strongest base.
So, why is C out of order in your series?
We are comparing #"CH"_3^-# and #"CH"_3"C≡C"^-#.
The difference is in the hybridization of the C atom.
In #"CH"_3^-#, the hybridization is #sp^3#.
In #"CH"_3"C≡C"^-#, the hybridization is #sp#.
An #sp^3# orbital has 25 % #s# character, while an #sp# orbital has 50 % #s# character.
#s# electrons are held more tightly to the nucleus, so an alkyne C atom is less basic than an alkane carbon.
The order of acidities (with #"p"K_"a"# values) is
#"CH"_4# (48) < #"NH"_3# (38) < #"CH"_3"C≡CH"# (25) < #"CH"_3"CH"_2"OH"# (16) < #"HF"# (3)
So the order of basicities is
#"CH"_3^-# > #"NH"_2^-# > #"CH"_3"C≡C"^-# > #"CH"_3"CH"_2"O"^-# > #"F"^-#
