We know: $\frac{\text { mass of } C }{\text { mass of glucose }}=\frac{72}{180}$

Given: $\% C =10.8=\frac{\text { mass of } C }{\text { mass of solution }} \times 100$

$\frac{10.8 \times 250}{100}=$ mass of $C \Rightarrow$ Mass of $C =27\,gm$

$\therefore$ mass of glucose $=67.5\,gm$

$\therefore$ moles of glucose $=0.375\,moles$

Mass of solvent $=250-67.5\,gm =182.5\,gm$

$\therefore$ Molality $=\frac{0.375}{0.1825}=2.055 \approx 2.06$