If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2\hat{u}\times 3\hat{v}$ is a unit vector for:

The magnitude of the cross product is given by $|2\hat{u}\times 3\hat{v}|=|2\hat{u}|\cdot |3\hat{v}|\cdot \sin \theta$.Since $\hat{u}$ and $\hat{v}$ are unit vectors, $|2\hat{u}|=2$ and $|3\hat{v}|=3$.We are given that $2\hat{u}\times 3\hat{v}$ is a unit vector, so its magnitude is $1$.
$$2\times 3\times \sin \theta =1$$
$$6\sin \theta =1⟹\sin \theta =1/6$$
Since $\theta$ is an acute angle, there is exactly one value of $\theta ={\sin }^{-1}(1/6)$ in $(0,\pi /2)$ that satisfies this.