Mole fraction of ethanol in ethanol water mixture is 0.25. Hence, the percentage concentration of ethanol by weight of mixture is

$\begin{array}{l}{\mathrm{\chi }}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.25\text{ (ethanol) }\left(\text{ solute) }\left({\mathrm{Mw}}_2=46 \mathrm{g}\right)\right.\\ {\mathrm{\chi }}_1=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.75\text{ (water) (solvent) }\left({\mathrm{Mw}}_1=18\mathrm{g}\right)\end{array}$

$\begin{array}{l}\frac{{\mathrm{\chi }}_1}{{\mathrm{\chi }}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{3}{1}\Rightarrow \frac{{\mathrm{W}}_1\times {\mathrm{Mw}}_2}{{\mathrm{Mw}}_1\times {\mathrm{W}}_2}=\frac{3}{1}\\ \frac{{\mathrm{W}}_1\times 46}{18\times {\mathrm{W}}_2}=3\\ \frac{{\mathrm{W}}_1}{{\mathrm{W}}_2}=1.17\\ \therefore \frac{{\mathrm{W}}_2}{{\mathrm{W}}_1+{\mathrm{W}}_2}=0.46\end{array}$