Vapor pressure depends on the strength of intermolecular forces (mainly London dispersion forces for these nonpolar/weakly polar molecules). Weaker intermolecular attractions → more molecules escape to the gas phase → higher vapor pressure.

Dispersion forces increase with molecular size/polarizability (roughly with molar mass and number of electrons):

• $\ce{CHCl3}$: $M \approx 119$ g/mol
• $\ce{CFCl3}$: $M \approx 137$ g/mol, but F is small and only weakly polarizable, and this molecule has the smallest surface area/polarizability among the heavier halogens present
• $\ce{CCl4}$: $M \approx 154$ g/mol
• $\ce{CBrCl3}$: $M \approx 198$ g/mol (largest, most polarizable, strongest dispersion forces)

Comparing the four, $\ce{CBrCl3}$ has the largest, most polarizable electron cloud (heaviest halogens), giving the strongest dispersion forces and lowest vapor pressure. $\ce{CFCl3}$ (Freon-11) contains a small, weakly polarizable F atom in place of one Cl, reducing the overall polarizability/surface area compared to $\ce{CCl4}$ or $\ce{CBrCl3}$, and it also lacks the H-bond-like dipole enhancement seen in $\ce{CHCl3}$. Experimentally, $\ce{CFCl3}$ (bp ≈ 24 °C) is the most volatile of the four, giving it the highest vapor pressure at 25 °C.

Thus the correct order of vapor pressure is $\ce{CFCl3} > \ce{CHCl3} > \ce{CCl4} > \ce{CBrCl3}$, matching the trend of increasing molar mass/polarizability with decreasing vapor pressure.