Solution:

$\sin 20^{0} \sin 40^{0} \sin 60^{0} \sin 80^{0}$

$=\frac{\sqrt{3}}{2}\left[\sin 40^{\circ} \sin 80^{\circ}\right] \sin 20^{\circ}$

$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}-\cos 120^{\circ}\right] \sin 20^{\circ}$

$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}-\cos \left(180^{\circ}-60\right)\right] \sin 20^{\circ}$

$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}+\cos 60^{\circ}\right] \sin 20^{\circ}$

$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}+\frac{1}{2}\right] \sin 20^{\circ}$

$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ} \sin 20^{\circ}+\frac{1}{2} \sin 20^{\circ}\right]$

$=\frac{\sqrt{3}}{8}\left[\sin 60^{\circ}+\sin \left(-20^{\circ}\right)+\sin 20^{\circ}\right]$

$=\frac{\sqrt{3}}{8}\left[\sin 60^{\circ}+\sin 20^{\circ}-\sin 20^{\circ}\right]$

$=\frac{\sqrt{3}}{8} \frac{\sqrt{3}}{2}=\frac{3}{16}$