3. As shown in Figure 2, in rhombus $A B C D$, the distances from vertex $A$ to sides $B C$ and $C D$ are $A E$ and $A F$, both equal to $5$, and $E F=6$. Therefore, the side length of rhombus $A B C D$ is $\qquad$
3. $\frac{125}{24}$.

As shown in Figure 2, connect $A C$ and $E F$, intersecting at point $G$. It is easy to prove that $E F \perp A C, E G=G F$.
In the right triangle $\triangle A E G$, since $A E=5, E G=\frac{1}{2} E F=3$, therefore, $A G=4$.

In the right triangle $\triangle A E C$, since $A E^{2}=A G \cdot A C, \angle A C E=$ $\angle A E F$, therefore, $A C=\frac{25}{4}$, and $\triangle B A C \backsim \triangle A E F$.
Thus, $\frac{B A}{A E}=\frac{A C}{E F}$, which means $A B=\frac{5 \times \frac{25}{4}}{6}=\frac{125}{24}$.