Why is the ionization enthalpy of francium greater than that of cesium?
The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.

This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.