Mole fraction of glucose= 0.2 implies that

Amount of glucose, n₂ = 0.2 mol;    Amount of water, n₁ = 0.8 mol

Mass of water, $m_1=n_1{M}_1=\left(0.8\mathrm{mol}\right)\left(18\mathrm{g}{\mathrm{mol}}^{-1}\right)=14.4\mathrm{g}=14.4\times {10}^{-3}\mathrm{kg}$

Molality of solution, $m=\frac{n_2}{m_1}=\frac{0.2\mathrm{mol}}{14.4\times {10}^{-3}\mathrm{kg}}=14.18\mathrm{mol}{\mathrm{kg}}^{-1}$