\begin{array}{l}
\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}=(12 \times 2)+(1 \times 2)+(16 \times 4)+2(18)\\
=(24+2+64+36) g \mathrm{~mol}^{-1}\\
=126 \mathrm{gmol}^{-1}\\
\% \text { of } \mathrm{H}_2 \mathrm{O}=\frac{2(18)}{126} \times 100 \%\\
=\frac{36}{126} \times 100 \%=28.57 \%
\end{array}