There are a couple of things you need to know for this problem: 1) that \lim_{x\to 0}\frac{\sin x}{x}=1. You can refer to this for a geometric argument from scratch, or use the Rule of de l'Hospital if you've seen it.
2) the limit comparison test, stating that for two series \sum_{n=1}^\infty a_n and \sum_{n=1}^\infty b_n with postive terms, if \lim_{n\to \infty}\frac{a_n}{b_n}=c for c a finite positive number, then either both series converge, or both diverge. See for instance here for details.
3) That p-series of the form \sum_{n=1}^\infty \frac{1}{n^p} converges iff p>1. See here for details.
Now, we are looking at the case a_n=\sin^2(\frac{1}{n}) and b_n=\frac{1}{n^2}.
Then \lim_{n\to \infty} \frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\sin^2(\frac{1}{n})}{\frac{1}{n^2}}=\lim_{n\to \infty}(\frac{\sin(\frac{1}{n})}{\frac{1}{n}})^2.
Let t=\frac{1}{n}. Then \lim_{n\to \infty}t=0 so that \lim_{n\to\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}}=\lim_{t\to 0}\frac{\sin t}{t}=1 and thus \lim_{n\to \infty} \frac{a_n}{b_n}=1^2=1>0.
By the Limit Comparison Test, \sum_{n=1}^\infty a_n and \sum_{n=1}^\infty b_n behave similarly (in terms of convergence). But we know that \sum_{n=1}^\infty \frac{1}{n^2} is a convergent p-series (p>1), so that \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{\sin^2(\frac{1}{n})}{\frac{1}{n^2}} is convergent.
