I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.
Via the disk method,
\(\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx\)
Recall the half-angle identity for sine:
\(\sin^2t=\dfrac{1-\cos2t}2\)
\(\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx\)
\(=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}\)
\(=\dfrac{9\pi^2}2\)
