As we know
$\frac{{\mathrm{p}}^{\mathrm{o}}-\mathrm{p}}{{\mathrm{p}}^{\mathrm{o}}}={\mathrm{x}}_1=$ mole fraction of solute
The ratio $\frac{\left({\mathrm{p}}^{\mathrm{o}}-\mathrm{p}\right)}{{\mathrm{p}}^{\mathrm{o}}}$ is the relative lowering of vapour pressure, which is equal to 0.0125 here.
So ${\mathrm{X}}_1=0.0125$
The relation between the mole fraction and molality is
$\left(\frac{1}{{\mathrm{X}}_1}-1\right)=\frac{1000}{\mathrm{m}\times 18}$ (molecular weight of ${\mathrm{H}}_2\mathrm{O}=18$ )
or $\left(\frac{1}{0.0125}-1\right)=\frac{100}{\mathrm{m}\times 18}$ w
$\mathrm{m}=\mathrm{0.70\; m}$