Suggestion: Make a change of variable $t \mapsto t/x^2$ and recall that $\sin(u)/u \to 1$ as $u \downarrow 0$.
Elaborating:
$ \frac{{\int_{x^2 }^{3x^2 } {t\sin (\frac{2}{t})dt} }}{{x^2 }} = \frac{{\int_1^3 {x^2 t\sin (\frac{2}{{x^2 t}})x^2 dt} }}{{x^2 }} = 2\int_1^3 {\frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg)dt} . $ Noting that $ \frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg) = \frac{{\sin (\frac{2}{{x^2 t}})}}{{2/(x^2 t)}} \to 1 $ uniformly for $t \in [1,3]$, it thus follows that the desired limit is equal to $2(3-1)$, that is to $4$.
EDIT (details concerning the uniform convergence mentioned above): Let $\varepsilon > 0$. Then $1-\varepsilon \leq \sin(u)/u \leq 1$ for all sufficiently small $u > 0$. Thus $ 1 - \varepsilon \le \frac{{\sin (\frac{2}{{x^2 t}})}}{{2/(x^2 t)}} \le 1 $ for all sufficiently large $x > 0$, uniformly in $t \in [1,3]$, since $0 < 2/(x^2 t) \leq 2/x^2 \to 0$. Hence $ 2\int_1^3 {(1 - \varepsilon )dt} \leq 2\int_1^3 {\frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg)dt} \leq 2\int_1^3 {1dt} $ for all sufficiently large $x$, implying that the limit, as $x \to \infty$, is $4$.