Answer
Converges to zero.
Work Step by Step
Sine is in the interval $[-1,1]$.
Its square is nonnegative and less (or equal to) 1.
$0\leq\sin^{2}n\leq 1$
$\displaystyle \frac{1}{2^{n}}$ is positive and multiplying the compound inequality with it will not change the inequality direction.
$0\displaystyle \leq\frac{\sin^{2}n}{2^{n}}\leq\frac{1}{2^{n}}$
Both $\{0\}$ and $\displaystyle \{\frac{1}{2^{n}}\}$ converge to zero. Applying the sandwich theorem, $\displaystyle \{\frac{\sin^{2}n}{2^{n}}\}$ also converges to zero.
