If $3 \sin 2 \theta=2 \sin 3 \theta$ and $0 < \theta < \pi$, then the value of $\sin \theta$ is equal to

Given the equation $3 \sin 2\theta = 2 \sin 3\theta$, apply the double and triple angle formulas to obtain $6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin^3 \theta$.

Divide both sides by $\sin \theta$ (since $\sin \theta \neq 0$ for $0 < \theta < \pi$), yielding $6 \cos \theta = 6 - 8 \sin^2 \theta$.

Substitute $\sin^2 \theta = 1 - \cos^2 \theta$ to form a quadratic in $\cos \theta$: $6 \cos \theta = 6 - 8(1 - \cos^2 \theta)$, which simplifies to $4 \cos^2 \theta - 3 \cos \theta - 1 = 0$.

Factor as $(4\cos\theta + 1)(\cos\theta - 1) = 0$, giving $\cos\theta = -\frac{1}{4}$ or $\cos\theta = 1$. Only $\cos\theta = -\frac{1}{4}$ lies within the domain $0 < \theta < \pi$.

Using $\sin^2 \theta = 1 - \cos^2 \theta$, compute $\sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = \frac{15}{16}$. Since $\sin \theta > 0$ in this interval, $\sin \theta = \frac{\sqrt{15}}{4}$.

The correct option is D.