Answer
$Na-Cl$ distance is longer than $K-F$ distance.
Work Step by Step
The attraction between 2 opposite-charged ions can be described by Coulomb's law. Here the electrostatic potential energy of 2 interacting charged bodies is $$E_{el}=\frac{kQ_1Q_2}{d}$$
$k$: Coulomb's law constant
$Q_1$: the magnitude of charge of body 1
$Q_2$: the magnitude of charge of body 2
$d$: the distance between 2 bodies' centers
Therefore, the attraction between 2 opposite-charged ions inversely depends on the distance between 2 bodies. In other words, the farther one body is from the other, the weaker the attraction between them is.
That means the farther the ions are among one another in a compound, the weaker the attraction among them is, and as a result, the easier to separate them into gaseous ions.
So the farther the ions are from one another in a compound, the smaller the lattice energy of that compound.
The lattice energy of $NaCl (788 kJ/mol)$ is smaller than that of $KF (808 kJ/mol)$. So the ions of $NaCl$ are farther from one another than the ions of $KF$. In other words, $Na-Cl$ distance is longer than $K-F$ distance.
