Step 1: Frame the comparison.
Acidity here means how easily $X\!-\!CH_2OH$ gives up its proton. Anything that steadies the resulting $RO^-$ makes the acid stronger.
Step 2: Start with methanol.
The $CH_3$ group pushes electron density toward the oxygen, which makes $CH_3O^-$ less stable. So plain methanol is the weakest acid of the set.
Step 3: Add the halogen pull.
A halogen on the carbon drains electron density away and spreads the negative charge, steadying the anion. So all three halo-alcohols beat methanol.
Step 4: Order the halogens in the gas phase.
Here size and polarizability help share the charge. Fluorine's strong electron withdrawal puts $FCH_2OH$ on top, and bromine's bigger, softer cloud handles the charge better than chlorine, so Br comes before Cl.
Step 5: Answer.
The order is \[ \boxed{FCH_2OH > BrCH_2OH > ClCH_2OH > CH_3OH} \]