2. In Rt $\triangle A B C$, $\angle C=90^{\circ}, B C=5$, and the other two sides $A B$ and $A C$ are of integer lengths. Extend $A C$ to $D$, and make $\angle A B D=90^{\circ}$ to get $\triangle A B D$. Then the circumradius $R$ of $\triangle A B D$ is $\qquad$
$2.7 \frac{1}{24}$.
By the Pythagorean theorem, we get $A B^{2}-A C^{2}=25$, which means $(A B+A C)(A B-A C)=25$.
Since the positive integer factors of 25 are $1,5,25$, we have
$$
\left\{\begin{array}{l}
A B+A C=25, \\
A B-A C=1 .
\end{array}\right.
$$

Solving, we get $A B=13, A C=12$.
By the projection theorem, we get
$$
A D=\frac{A B^{2}}{A C}=14 \frac{1}{12} \text {. }
$$

Therefore, $R=\frac{1}{2} A D=7 \frac{1}{24}$.