taking assumption that $|\alpha |>1$
$P(|x|<1) = \int_{-1}^{1}\frac{1}{2\alpha }dx = $$\frac{1}{\alpha }$
$P(|x|>1) = \int_{-\alpha }^{-1}\frac{1}{2\alpha }dx+\int_{1}^{\alpha }\frac{1}{2\alpha }dx=$$\frac{-1+\alpha }{2\alpha } +\frac{\alpha -1}{2\alpha }= \frac{\alpha -1}{\alpha }$
$\Rightarrow \frac{\alpha -1}{\alpha }=\frac{1}{\alpha }$
$\Rightarrow \alpha =2$