Use the identity ${\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x=1-\frac{{\sin }^{2}2x}{2}$.

Then $I={\int }_0^{20\pi }\left(1-\frac{{\sin }^{2}2x}{2}\right)dx=20\pi -\frac{1}{2}{\int }_0^{20\pi }{\sin }^{2}2xdx$.

By periodicity, ${\int }_0^{20\pi }{\sin }^{2}2xdx=40\cdot {\int }_0^{\pi /2}{\sin }^{2}2xdx=40\cdot \frac{\pi }{4}=10\pi$.

$\therefore I=20\pi -\frac{1}{2}(10\pi )=15\pi$.