3. $\frac{2 \pi}{7}$.

As shown in Figure 1, take a point $D$ on the extension of $C A$ such that $A D=A I$. Then
$$\begin{array}{l} C D=C A+A D \\ =C A+A I=B C . \end{array}$$

By the given conditions,
$$\begin{array}{l} \angle C D I=\angle C B I \\ =\angle A B I . \end{array}$$

Since $A D=A I$, we have
$$\angle C A I=2 \angle A D I=2 \angle A B I=\angle A B C \text {. }$$

Thus, $\angle C A B=2 \angle A B C$.
Similarly, $\angle A B C=2 \angle A C B$.
Then $\pi=\angle A B C+\angle C A B+\angle A C B=\frac{7}{2} \angle A B C$
$$\Rightarrow \angle A B C=\frac{2 \pi}{7} \text {. }$$