Denote $a = BC$, $b = AC$, $c = AB$, so $b = 2$, $c = 1$. By the law of sines, we have $\frac{\sin B}{\sin C} = \frac{b}{c} = 2$, namely, $\sin B = 2 \sin C$. And since $\cos B = 1 - \sin C$, there is
$$(2 \sin C)^2 + (1 - \sin C)^2 = \sin^2 B + \cos^2 B = 1,$$
and simplifying it gives $5\sin^2 C - 2\sin C = 0$. And because $\sin C \neq 0$, $\sin C = \frac{2}{5}$.
And then we have $\cos B = 1 - \sin C = \frac{3}{5}$.
By the law of cosines, we have $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$, and thus $\frac{3}{5} = \frac{a^2 - 3}{2a}$, i.e.,
$$a^2 - \frac{6}{5}a - 3 = 0.$$
Since $a > 0$, we get the solution $a = \frac{3 + 2\sqrt{21}}{5}$, namely the length of side $BC$ is $\frac{3 + 2\sqrt{21}}{5}$. $\square$