3. (3 points) There are two cylinders with a volume ratio of 5:8, and the lateral surfaces of these cylinders unfold into the same rectangle. If the length and width of this rectangle are both increased by 6, its area increases by 114. What is the area of this rectangle $\qquad$ .
【Analysis】The lateral development is a rectangle of the same size, indicating that this rectangle can be wrapped horizontally to form a cuboid and vertically to form another cuboid, with a volume ratio of 5:8. As shown in the figure, the area of the shaded part is 114, so $(a+b)$ is $(114 - 6 \times 6) \div 6 = 13$. According to the volume ratio of 5:8, we have: $\frac{\left(\frac{a}{2 \pi}\right)^{2} \cdot \pi \cdot b}{\left(\frac{b}{2 \pi}\right)^{2} \cdot \pi \cdot a} = \frac{5}{8}$, which simplifies to $\frac{\frac{a^{2}}{4 \pi} \cdot b}{\frac{b^{2}}{4 \pi} \cdot a} = \frac{5}{8}$, and further simplifies to $\frac{a}{b} = \frac{5}{8}$. Given that $a + b = 13$, $a$ and $b$ are 8 and 5 respectively, and the area is $5 \times 8 = 40$. Based on this, the solution is as follows.
【Solution】Let the length and width of the rectangle be $a$ and $b$, respectively, then $a + b = (114 - 6 \times 6) \div 6 = 13$,
According to the volume ratio of 5:8, we have: $\frac{\left(\frac{a}{2 \pi}\right)^{2} \cdot \pi \cdot b}{\left(\frac{b}{2 \pi}\right)^{2} \cdot \pi \cdot a} = \frac{5}{8}$,
Simplifying to $\frac{\frac{a^{2}}{4 \pi} \cdot b}{\frac{b^{2}}{4 \pi} \cdot a} = \frac{5}{8}$,
Further simplifying to $\frac{a}{b} = \frac{5}{8}$,
Given that $a + b = 13$,
So $a$ and $b$ are 8 and 5 respectively, and the area is $5 \times 8 = 40$,
Answer: The area of this rectangle is 40.
Therefore, the answer is: 40.