3. Given $I$ is the incenter of $\triangle A B C$, $\angle A B C=2 \angle A C B, A B=C I$. Then $\angle B I C=$ $\qquad$
3. $120^{\circ}$.

As shown in Figure 2, connect $B I$ and extend it to intersect the circumcircle of $\triangle A B C$ at point $D$, and intersect $A C$ at point $E$; extend $C I$ to intersect $A B$ at point $F$; connect $A D$ and $C D$.
It is easy to see that $B E=C E$.
By symmetry, quadrilateral $A B C D$ is an isosceles trapezoid.
Then $C I=A B=C D$
$\Rightarrow \angle E I C=\angle B D C=\angle F A E$
$\Rightarrow A, E, I, F$ are concyclic
$\Rightarrow \angle B A C=180^{\circ}-\angle B I C$.
By the properties of the incenter,
$\angle B I C=90^{\circ}+\frac{1}{2} \angle A B C$.
Combining the two equations, we get $\angle B I C=120^{\circ}$.