$m=$ molality, $X_{\text {solute }}=$ Mole fraction of solute, $X_{\text {solvent }}=$ Mole fraction of solvent, $M_{\text {solvent }}=$ Molar mass of solvent

Here solute is methyl alcohol $(C{H_3}OH)$ and solvent is water $(H_{2} \mathrm { O } )$ Here water is solvent because question says solution is aqueous

Assume mole fraction of methyl alcohol $=x$

So mole fraction of water $=1-X$

and given molality $(\mathrm{m})=5.2$

$\mathrm{M}_{\text {solvent }}=18$ (as molar mass of $\mathrm{H}_{2} \mathrm{O}=18$ )

$\therefore 5.2=\frac{X}{1-X} \times \frac{1000}{18}$

by solving this we get,

$X=0.086$