Given :
$P_s=\big(\large\frac{98}{100}\big)$$P^0$
We know $\large\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times W}$
$\Rightarrow \large\frac{w}{m\times W}$$\times 1000\times \large\frac{M}{1000}$
As molality =$\large\frac{\text{number of moles of the solute}}{\text{Mass of the solvent in kilograms}}$
$\Rightarrow \large\frac{w\times 1000}{m\times W}$
$\therefore \large\frac{P^0-(98/100)P^0}{(98/100)P^0}$$=Molality \times \large\frac{18}{1000}$
$\therefore$ Molality =$\large\frac{2P^0}{100\times \large\frac{98}{100}P^0}$$\times \large\frac{1000}{18}$
$\Rightarrow 1.133$