$$
\begin{array}{l}
A=\{x|| x-2 \mid<a\}, \\
B=\left\{x \mid x^{2}-2 x-3<0\right\} .
\end{array}
$$

If $B \subseteq A$, then the range of real number $a$ is $\qquad$
$$
-1 . a \geqslant 3 \text {. }
$$

From the problem, we know that $A=\{x \mid 2-a<x<2+a\}$,
$$
B=\{x \mid-1<x<3\} \text {. }
$$

Since $B \subseteq A$, then $2-a \leqslant-1$, and $3 \leqslant 2+a$. Therefore, $a \geqslant 3$.