The hydrogen ion transfer is onto the methylamine, giving methylammonium ion as the product. The other product is fluoride ion.
HF(aq) + CH3NH2(aq) ? CH3NH3+(aq) + F–(aq)
We need to find Ka values using the Table of Acid Ionization Constants
Ka(acid) = 6.6×10–4
Ka(conjugate acid) = 2.4×10–11
So Kc = Ka(acid)/Ka(conjugate acid) = (6.6×10–4)/(2.4×10–11) = 2.8×107 > 100