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