B. As shown in Figure $5, \odot O$ has a radius of $20, A$ is a point on $\odot O$. A rectangle $O B A C$ is constructed with $O A$ as the diagonal, and $O C=$ 12. Extend $B C$ to intersect $\odot O$ at points $D$ and $E$, then $C E-B D$ $=$ . $\qquad$
B. $\frac{28}{5}$.

Let the midpoint of $D E$ be $M$, and connect $O M$. Then $O M \perp D E$.

Since $O B=\sqrt{20^{2}-12^{2}}=16$, we have,
$$
\begin{array}{l}
O M=\frac{O B \cdot O C}{B C}=\frac{16 \times 12}{20}=\frac{48}{5}, \\
C M=\sqrt{O C^{2}-O M^{2}}=\frac{36}{5}, B M=\frac{64}{5} .
\end{array}
$$

Therefore, $C E-B D=(E M-C M)-(D M-B M)$
$$
=B M-C M=\frac{64}{5}-\frac{36}{5}=\frac{28}{5} \text {. }
$$