imc

By Pythagorean Theorem in three dimensions, the distance $XY$ is $\sqrt{4^2+4^2+10^2}=2\sqrt{33}$. Let the length of the segment $XY$ that is inside the cube with side length $3$ be $x$. By similar triangles, $\frac{x}{3}=\frac{2\sqrt{33}}{10}$, giving $x=\boxed{\frac{3\sqrt{33}}{5}}skr-$.