Solution:

Let $X$ and $Y$ be the points where the incircle touches $AB$ and $BC$, respectively. Then $AXI$ and $AYB$ are similar right triangles. Since $I$ is the incenter, we have $IX = IY = 2$. Using the Pythagorean theorem on triangle $AXI$, we find $AX = \sqrt{5}$. By similarity, $AY / AX = BY / IX$. Plugging in the numbers given, $5 / \sqrt{5} = BY / 2$, so $BY = 2\sqrt{5}$. $Y$ is the midpoint of $BC$, so $BC = 4\sqrt{5}$.