In the replacement reaction:

The reaction will be most favourable if $\text{M}$ happens to be:

This is a Swarts-type reaction. For the reaction to be most favourable, the metal fluoride ($\text{MF}$) should have lower lattice energy and more ionic character, which increases the reactivity of the fluoride ion. Down Group 1, as the size of the cation increases (${\text{Li}}^{+}<{\text{Na}}^{+}<{\text{K}}^{+}<{\text{Rb}}^{+}$), the lattice energy of the fluoride decreases. Rubidium fluoride ($\text{RbF}$) would be the most reactive among the given options.