jmc

Let the cone have height $h$ and radius $r$, so its volume is $$\frac{1}{3}\pi r^{2}h.$$When the cone is filled with water, the amount of water in the cone forms a smaller cone that is similar to the original cone. This smaller cone has height $\frac{3}{4}h$, and by similar triangles, radius $\frac{3}{4}r$. So, the smaller cone has volume $$\frac{1}{3}\pi {\left(\frac{3}{4}r\right)}^2\left(\frac{3}{4}h\right)=\frac{1}{3}\pi \cdot \frac{3^3}{4^3}{r}^{2}h.$$Hence the ratio of the volume of the water-filled cone to the original cone is $$\frac{3^3}{4^3}=\frac{27}{64}=0.421875,$$which, as a percentage, is $\boxed{42.1875}\%$.