$$\begin{gathered} {\mathrm{X}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}=0.1 \\ \mathrm{If} \mathrm{the} \mathrm{total} \mathrm{mole} \mathrm{is} 1 \mathrm{mol}, \mathrm{then} \mathrm{the} \mathrm{moles} \mathrm{of} \mathrm{glucose} \mathrm{and} \\ \mathrm{water} \mathrm{will} \mathrm{each} \mathrm{be} 0.1 \mathrm{and} 0.9, \mathrm{respectively}. \\ \mathrm{The} \mathrm{mass} \mathrm{percent} \mathrm{of} \mathrm{water} \mathrm{is}=\frac{(0.9 \mathrm{x} 18)}{(0.1 \mathrm{x} 180 + 0.9 \mathrm{x} 18)} \mathrm{x} 100=47.36 \\ \approx 47 \end{gathered}$$

Therefore, the correct option is A 47.00