The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:

NaCl dissociates in water: $\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$. One formula unit produces two particles in solution. Van't Hoff factor $i = \dfrac{\text{actual moles of particles}}{\text{moles of solute formula units}}$. For complete dissociation of NaCl, $i = 2$. In reality, ion pairing slightly reduces $i$ below the theoretical value, especially at higher concentrations. At dilute $0.01$ M the assumption of complete dissociation is reasonable, and the experimental $i$ is close to $1.9$. The question asks for the theoretical value assuming complete dissociation, so 2 is the cleanest answer. Compare with $\mathrm{CaCl_2}$: dissociates into 3 ions, so $i = 3$ (option D) — common trap.