4. If the surface area of a cube $X$ is equal to that of a regular tetrahedron $Y$, then the ratio of their volumes $\frac{V_{X}}{V_{Y}}=$ $\qquad$ .
4. $\sqrt[4]{3}$.

Let the surface area be 12.
Then the area of each face of the cube is 2, and its side length is $\sqrt{2}$. Thus, $V_{x}=2^{\frac{3}{2}}$.

Given that the area of each face of the regular tetrahedron is 3, let its side length be a. Then
$$
\frac{\sqrt{3}}{4} a^{2}=3 \Rightarrow a=2 \times 3^{\frac{1}{4}} .
$$

Thus, $V_{Y}=2^{\frac{3}{2}} \times 3^{-\frac{1}{4}}$.
Therefore, $\frac{V_{X}}{V_{Y}}=3^{\frac{1}{4}}=\sqrt[4]{3}$.