$∆{\mathrm{T}}_{\mathit{t}}=\left(273.15-269.15\right)\mathrm{K}=4\mathrm{K}$
Molar mass of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$
$=\left(12\times 12\right)+\left(22\times 1\right)+\left(11\times 16\right)$
$=342\mathrm{g}{\mathrm{mol}}^{-1}$
$10\%$ solution of sucrose in water means $10\mathrm{g}$ of sucrose is present in $\left(100-10\right)\mathrm{g}$ of water.

$=0.0292\mathrm{mol}$
Therefore, molality of the solution
$=\frac{0.0292\times 1000}{90}=0.3244\mathrm{mol}{\mathrm{kg}}^{-1}$
We know that $∆{\mathrm{T}}_{\mathit{t}}={\mathrm{K}}_{\mathit{f}}\times \mathit{m}$
$\Rightarrow {\mathrm{K}}_{\mathit{f}}=\frac{∆{\mathrm{T}}_{\mathit{t}}}{\mathit{m}}=\frac{4}{0.3244}$ $=12.33\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$
Molar mass of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$
$=\left(6\times 12\right)+\left(12\times 1\right)+\left(6\times 16\right)$ $=180\mathrm{g}{\mathrm{mol}}^{-1}$
$10\%$ solution of glucose in water means $10\mathrm{g}$ of glucose is present in $\left(100-10\right)\mathrm{g}$ of water.
$=\frac{10}{180}=0.0555\mathrm{mol}$
Therefore,molality of the solution
$=\frac{0.0555\times 1000}{90}$
$=0.6166\mathrm{mol}{\mathrm{kg}}^{-1}$

$\Rightarrow ∆{\mathrm{T}}_{\mathit{t}}=12.33\times 0.6166=7.60\mathrm{K}$
So, the freezing point of $10\%$ glucose solution in water is $\left(273.15-7.60\right)\mathrm{K}=265.55\mathrm{K}$