[Solution] If the radius of the wheel is $r$, then the distance $D$ and the number of wheel rotations $N$ satisfy $D=N \cdot$ $2 \pi r$.

Let $N_{1}, N_{2}$ be the number of rotations for the outbound and return trips, and $r_{1}, r_{2}$ be the radii of the wheels for the outbound and return trips, respectively. According to the problem, we have
$2 \pi r_{1} N_{1}=2 \pi r_{2} N_{2}$, then $r_{2}: r_{1}=N_{1}: N_{2}=450: 440$.
Given $r_{1}=15$, then $r_{2}=\frac{15 \cdot 45}{44}$,
$\therefore \quad r_{2}-r_{1}=15 \cdot\left(\frac{45}{44}-1\right)=\frac{15}{44} \approx 0.34$.

Therefore, the answer is (B).
3.36 In a one-way trip, the distance read from the instrument is 450 miles. On the return trip, using snow tires to travel the same distance, the reading is 440 miles. If the original wheel radius is 15 inches, find the increase in the wheel radius (to the nearest hundredth of an inch).
(A) 0.33.
(B) 0.34.
(C) 0.35.
(D) 0.38.
(E) 0.66.
(21st American High School Mathematics Examination, 1970)