Answer
$e^{-(1 / \ln 2)} $; Converges
Work Step by Step
Suppose $\dfrac{1}{x}=y; \dfrac{-1}{x^2}dx=dy$
$ \int_{0}^{\ln 2} x^{-2} (e^{-1 / x}) d x = \int_{\infty}^{(1 / \ln 2)} \dfrac{y^{2} e^{-y} d y}{-y^{3}}\\=\int_{1 / \ln 2}^{\infty} e^{-y} d y\\
=\lim\limits_{b \to \infty}\int_{1 / \ln 2}^{b} e^{-y} d y$
Now, $=\lim\limits_{b \to \infty} [-e^{-y}]|_{1 / \ln 2}^{b}$
or, $=\lim\limits_{b \to \infty} [-e^{-b}-(-e^{-1 / \ln 2})]$
or, $=0+e^{-1 / \ln 2}$
or, $=e^{-(1 / \ln 2)} $
So, the integral converges.