Answer: 2

Solution:

Write $a = \sqrt{2014},$ so the equation is $ax^3 - (2a^2 + 1)x^2 + 2 = 0.$ It factors as $$\left(ax - 1\right)\left(x^2 - 2ax - 2\right) = 0,$$ as expanding confirms. So one root is $\frac{1}{a},$ and the other two are $a \pm \sqrt{a^2 + 2},$ with product $-2$ and sum $2a.$

Since $a - \sqrt{a^2+2} \lt 0$ $\lt \frac{1}{a} \lt a + \sqrt{a^2+2},$ the middle root is $x_2 = \frac{1}{a},$ and $x_1 + x_3 = 2a.$ Therefore $x_2(x_1 + x_3) = \frac{1}{a} \cdot 2a = 2.$