Easy, Isn't It? ;)

Well I don't think there's anything wrong in this question,

Talking about the solution, we have:

sin20.sin40.sin60.sin80

= $\sqrt{3}$ . $\frac{1}{2}$ . sin 20 .sin 40 .sin 80

= $\sqrt{3}$ . $\frac{1}{2}$ . sin 20 .sin (60-20) .sin (60+20)

= $\sqrt{3}$ . $\frac{1}{2}$ . sin 20 .[ $sin^{2}60$ - $sin^{2}20$ ]

(we know that sin (A-B) .sin (A+B) = $sin^{2}A$ - $sin^{2}B$ )

= $\sqrt{3}$ . $\frac{1}{2}$ . sin 20 . [ \frac{3}{4} - \(sin^{2}20 ]

= $\sqrt{3}$ . $\frac{1}{2}$ . \frac{1}{4} . sin 20 . [ 3- 4\(sin^{2}20 ]

= $\sqrt{3}$ . $\frac{1}{2}$ . \frac{1}{4} . [ 3sin20- 4\(sin^{3}20 ]

= we know that 3sinA - 4 $sin^{3}20$ =sin 3A

= $\sqrt{3}$ . $\frac{1}{2}$ . $\frac{1}{4}$ . (sin 3(20))

= $\sqrt{3}$ . $\frac{1}{2}$ . $\frac{1}{4}$ . sin 60

= $\sqrt{3}$ . $\frac{1}{8}$ . $\sqrt{3}$ . $\frac{1}{2}$

= $\frac{3}{16}$

Thus the question as well as the answer is totally correct!!