Answer: A

Solution:

The distance between $X$ and $Y$ with respect to the $z$-axis is $$1 + 2 + 3 + 4 = 10.$$

Both the distances along the $x$ and $y$-axes are $4.$

Then $$XY = \sqrt{4^2 + 4^2 + 10^2} = 2\sqrt{33}.$$

Using coordinates $X=(0,0,10)$ and $Y=(4,4,0)$, the line meets the top and bottom of the side-$3$ cube at $(\frac65,\frac65,7)$ and $(\frac{12}5,\frac{12}5,4)$. Both points lie inside those square faces, so the portion inside this cube really does have vertical change $3$.

Let the desired length be $x.$ Then using similar triangles, we have that $$\dfrac{x}{3} = \dfrac{2\sqrt{33}}{10}$$$$x = \dfrac{3\sqrt{33}}{5}.$$

Thus, A is the correct answer.