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 cross product of two vectors $$\vec{A}$$ and $$\vec{B}$$ satisfies
$$|\vec{A}\times\vec{B}| = |\vec{A}|\,|\vec{B}|\,\sin\theta$$, where $$\theta$$ is the angle between them.

Here $$\vec{A} = 2\hat{u}$$ and $$\vec{B} = 3\hat{v}$$. Since $$\hat{u}$$ and $$\hat{v}$$ are unit vectors,
$$|\vec{A}| = 2 \quad\text{and}\quad |\vec{B}| = 3$$.

Therefore
$$|2\hat{u}\times 3\hat{v}| = (2)(3)\sin\theta = 6\sin\theta.\tag{1}$$

The vector $$2\hat{u}\times 3\hat{v}$$ will be a unit vector if and only if its magnitude equals 1. Using (1):
$$6\sin\theta = 1 \;\Longrightarrow\; \sin\theta = \frac{1}{6}.\tag{2}$$

Given that $$\theta$$ is acute, $$0 \lt \theta \lt \frac{\pi}{2}$$. In this interval the sine function is one-to-one, so equation (2) has exactly one solution:
$$\theta = \sin^{-1}\!\left(\dfrac{1}{6}\right).$$

Thus there is exactly one acute angle $$\theta$$ for which $$2\hat{u}\times 3\hat{v}$$ is a unit vector.

Option D which is: exactly one value of $$\theta$$