Aqueous $NaOH$ solution is labelled as $10\%$ by weight mole fraction of the solute in it is

Given, $w/w\%\text{ of }NaOH=10\%$

Let the weight of the solution is 100 g

Thus, weight of NaOH = 10 g

Moles of NaOH = $\frac{\text{ mass }}{\text{ molar mass }}=0.25\text{ moles }$

Mass of $H_{2}O=100–10=90g$

Moles of $H_{2}O=5moles$

Mole fraction of NaOH = $\frac{\text{ Moles of }NaOH}{\text{ Total moles }}=\frac{0.25}{5.25}=0.0476$

Therefore, Option B is correct.