Correct Answer: $\frac{\sqrt{15}-1}{4}$

Solution : Given that
$\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$
⇒ $\sin \theta=2 \cos ^2 \theta-\frac{1}{4}$
⇒ $\sin \theta=2-2 \sin ^2 \theta-\frac{1}{4}$
⇒ $2 \sin ^2 \theta + \sin \theta=\frac{7}{4}$
Let $\sin\theta=x$
Hence, the Expression becomes:
$8x^2+4x-7=0$
Comparing with $ax^2+bx+c=0$
$a=8, b=4,$ and $c=-7$
Which gives us
$x=\frac{-4\pm \sqrt{4^2-4×8×(-7)}}{2×8}$
⇒$x=\frac{-1\pm \sqrt{15}}{4}$
⇒ $x =\frac{\sqrt{15}-1}{4}$
⇒ $\sin\theta = \frac{\sqrt{15}-1}{4}$
Hence, the correct answer is $\frac{\sqrt{15}-1}{4}$.