If $\vec{A}\times \vec{B}=\vec{B}\times \vec{A}$, then the angle between $\vec{A}$ and $\vec{B}$ is:

We know that the vector cross product is anti-commutative:
$$\vec{A}\times \vec{B}=-(\vec{B}\times \vec{A})$$
Given:
$$\vec{A}\times \vec{B}=\vec{B}\times \vec{A}$$
Substitute the anti-commutative property:
$$\vec{A}\times \vec{B}=-(\vec{A}\times \vec{B})$$
$$2(\vec{A}\times \vec{B})=0⟹\vec{A}\times \vec{B}=0$$
The cross product is zero when the vectors are parallel or anti-parallel. Thus, the angle $\theta$ is $0$ or $\pi$. Among the given options, $\pi$ is correct.