We assume the side length of the cube is $1$. The side length of the tetrahedron is $\sqrt2$, so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$. The surface area of the cube is $6\times1\times1=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{\sqrt3}$.
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