$M_X\,=\,\frac {3}{4}M_Y$    .... $(i)$

The relative lowering of vapour pressure of two solution is

${\left( {\frac{{\Delta P}}{P}} \right)_X}\, = \,m{\left( {\frac{{\Delta P}}{P}} \right)_Y}$

But, the relative loering of vaoour pressure of solutin is directly proportional to the mole fraction of solution.

Given $5\,molal$ solution , mans $5\,molal$ of solute are dissolved in $1\,kg$ (or $1000\,g$) of solvent.

The number of moles of solvent $=\,\frac {1000\,g}{M}$

The mole fraction of solution $=\,\frac {5}{1000/M}$

                                              $=M\times \,\frac {5}{1000}$

hence $M_X\times \frac {5}{1000}$ $=\,m\times M_Y\times \frac {5}{1000}$   .... $(ii)$

Substitute equation $(i)$ in equation $(ii)$

$\frac {3}{4}\times M_Y\times \frac {5}{1000}$ $=\,m\times M_Y\times \frac {5}{1000}$

$m=\frac {3}{4}$