The answer is
$$\int_0^\infty dx\,\frac{e^{-x^2}\sin^2 x}{x^2}=\frac{\pi}{2}\,{\rm erf}(1)-\frac{\sqrt{\pi}}{2e}(e-1).$$
Consider
$$I(a)=\int_0^\infty dx\,\frac{e^{-ax^2}\sin^2 x}{x^2}$$
for which
$$-I'(a)=\int_0^\infty dx\,e^{-ax^2}\sin^2 x.$$
This last integral can be done by expanding sine in terms of exponentials and completing the square in each term. The result is
$$-I'(a)=\frac{\sqrt{\pi}}{16a}e^{-1/a}\left(e^{1/a}-1\right)$$
which can be anti-differentiated (I used Mathematica) in terms of the error function. As $a\to\infty$, $I(a)\to 0$, setting the constant of integration. Plugging in $a=1$ gives your answer.