[Solution] We first provide the following figure:
Since the projection length of the meter stick is $2 \mathrm{~m}$, we have $A B$ $=5$, and $A D=5 \sqrt{5}$,
and $\triangle C E A \backsim \triangle D B A$, so we have
$\frac{C E}{A C}=\frac{B D}{A D}$, which means $\frac{r}{5-r}=\frac{10}{5 \sqrt{5}}$.
Thus, $\sqrt{5} r=10-2 r$, which gives
$$
r=\frac{10}{2+\sqrt{5}}=10 \sqrt{5}-20 \text {. }
$$

Therefore, the answer is $(E)$.
24.11 In the sunlight, a sphere is placed on a plane. At a certain moment, the farthest point of the sphere's shadow from the point of tangency between the sphere and the plane is $10 \, \text{m}$. At the same moment, a meter stick perpendicular to the plane casts a shadow of $2 \, \text{m}$. What is the radius of the sphere (unit: $\text{m}$, assuming the sunlight is parallel, and the meter stick is a line segment)?
(A) $\frac{5}{2}$.
(B) $9-4 \sqrt{5}$.
(C) $8 \sqrt{10}-23$.
(D) $6-\sqrt{15}$.
(E) $10 \sqrt{5}-20$.
(34th American High School Mathematics Examination, 1983)