Answer
$30.1\%$
Work Step by Step
Let the volume of the solution be 1 L.
Then, molarity= 10.0 M implies that 10.0 moles of NaOH is in 1 L of solution.
Mass of solute=$ 10.0\,mol\times\frac{39.997\,g}{1\,mol}=399.97\,g$
Mass of the solution= Density$\times$Volume
$=\frac{1.33\,g}{1\,cm^{3}}\times\frac{1000\,cm^{3}}{1\,L}\times1\,L=1330\,g$
Weight percent=$ \frac{\text{mass of solute}}{\text{mass of solution}}\times100\%$
$=\frac{399.97\,g}{1330\,g}\times100\%=30.1\%$
