3. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ with left and right foci at $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively, and an eccentricity of 2. If the line $y=2(x-c)$ intersects the hyperbola at points $A$ and $B$, and the incenter of $\triangle A F_{1} F_{2}$ and $\triangle B F_{1} F_{2}$ are $I_{1}$ and $I_{2}$, respectively, and $\left|I_{1} I_{2}\right|=2 \sqrt{5}$, then the value of $a$ is $\qquad$
3. 2 .

It is known that $I_{1} I_{2}$ is perpendicular to the $x$-axis and the foot of the perpendicular is the right vertex.
In the trapezoid $I_{1} I_{2} S Q$, we have $\sin \angle Q I_{1} I_{2}=\frac{2}{\sqrt{5}}$.
Thus, $2 a=\left|I_{1} I_{2}\right| \sin \angle Q I_{1} I_{2}=2 \sqrt{5} \times \frac{2}{\sqrt{5}}=4$ $\Rightarrow a=2$