Answer:
Twice as much.
Step-by-step explanation:
That's because the freezing point depression depends on the total number of solute particles.
C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)
0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).
That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.