Answer is (B)
The above equation can be converted in the form of identity of the form:
$(a+b)^3 = a^3 + b^3 + 3ab(a+b)$
Let $a = \sin^2\frac{\pi}{81}, b = \cos^2\frac{\pi}{81}$
Now,
$=(\sin^2\frac{\pi}{81}+\cos^2\frac{\pi}{81})^3 = (\sin^2\frac{\pi}{81})^3 + (\sin^2\frac{\pi}{81})^3 + 3sin^2\frac{\pi}{81}cos^2\frac{\pi}{81}\underbrace{(\sin^2\frac{\pi}{81}+\cos^2\frac{\pi}{81})}_\text{=1} = 1$
But the question has extra $-1$. So, answer = $1-1$ = 0