1. **State the problem:** Solve the equation $$ (2r^2) \sin^3(2a) = \frac{r^2}{4} $$ for $\sin(2a)$. 2. **Isolate terms:** Divide both sides by $r^2$ (assuming $r \neq 0$) to get $$ 2 \sin^3(2a) = \frac{1}{4} $$ 3. **Solve for $\sin^3(2a)$:** $$ \sin^3(2a) = \frac{1}{8} $$ 4. **Take the cube root:** $$ \sin(2a) = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} $$ 5. **Final answer:** The value is $$ \sin(2a) = \frac{1}{2} $$ This means $2a$ could be any angle where sine equals $\frac{1}{2}$, such as $2a = 30^\circ$ or $150^\circ$ (or in radians $\frac{\pi}{6}, \frac{5\pi}{6}$) plus integer multiples of $2\pi$.