$\mathrm{CaO}$ and $\mathrm{NaCl}$ have the same crystal structure and approximately the same ionic radii. If $U$ is the lattice energy of $\mathrm{NaCl}$, the approximate lattice energy of $\mathrm{CaO}$ is

$$ \text { Lattice energy }=-\frac{q_1 q_2}{r^2} $$ where $q_1$ and $q_2$ are charges on ions and $r$ is the distance between them. Since interionic distances in $\mathrm{CaO}$ and $\mathrm{NaCl}$ are similar, (larger cation has smaller anion and vice versa) therefore, $r$ is almost the same. Therefore, lattice energy depends only on charge. Since the magnitude of charge on $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ions is unity and that on $\mathrm{Ca}^{2+}$ and $\mathrm{O}^{2-}$ ions is 2 each, therefore, the lattice energy of $\mathrm{CaO}$ is four times the lattice energy of $\mathrm{NaCl}$, i.e., $4 U$.