1. Formula and molar mass of D-glucose

D-glucose has formula: $\mathrm{C_6H_{12}O_6}$

Molar mass: $6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180\ \text{g mol}^{-1}$

2. Carbon content in glucose

Mass of carbon in $1$ mole of glucose: $6 \times 12 = 72\ \text{g}$

So, mass fraction of carbon in glucose is: $\frac{72}{180} = 0.4$

Thus, $40\%$ of glucose by mass is carbon.

3. Carbon present in the given solution

The solution mass is $250\ \text{g}$ and it contains $10.8\%$ carbon by weight.

So, mass of carbon in solution is: $\frac{10.8}{100} \times 250 = 27\ \text{g}$

4. Mass of glucose in the solution

Since glucose is $40\%$ carbon by mass, $0.4 \times (\text{mass of glucose}) = 27$

Therefore, $\text{mass of glucose} = \frac{27}{0.4} = 67.5\ \text{g}$

5. Mass of solvent (water)

Total solution mass = $250\ \text{g}$

So, mass of water: $250 - 67.5 = 182.5\ \text{g} = 0.1825\ \text{kg}$

6. Moles of glucose

$n = \frac{67.5}{180} = 0.375\ \text{mol}$

7. Molality

Molality is: $m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.375}{0.1825}$

$m \approx 2.055 \approx 2.06$

8. Match with options

Nearest value is: $\boxed{2.06}$ So, the correct option is B.