The correct answer is $\boxed{\frac{2}{3}}$.
To evaluate the integral, we can use the following identity:
$$\int \sin^3 \theta \, d\theta = \frac{1}{4} \sin \theta – \frac{3}{8} \cos \theta + \frac{1}{12} \sin 3\theta + C$$
where $C$ is an arbitrary constant.
Substituting in the limits of integration, we get:
$$\begin{align}
\int_0^\pi \sin^3 \theta \, d\theta &= \left[ \frac{1}{4} \sin \theta – \frac{3}{8} \cos \theta + \frac{1}{12} \sin 3\theta \right]_0^\pi \
&= \frac{1}{4} – \frac{3}{8} + \frac{1}{12} – \left( \frac{0}{4} – \frac{3}{8} + \frac{0}{12} \right) \
&= \frac{1}{4} – \frac{3}{8} + \frac{1}{12} – \frac{1}{4} + \frac{3}{8} + \frac{0}{12} \
&= \frac{2}{3}
\end{align}$$
Therefore, the integral $\int_0^\pi \sin^3 \theta \, d\theta$ is given by $\boxed{\frac{2}{3}}$.
The other options are incorrect because they do not give the correct value of the integral.