3. There are three square pools, large, medium, and small, with inner side lengths of 6 meters, 3 meters, and 2 meters, respectively. Two piles of gravel are submerged in the medium and small pools, causing the water levels to rise by 6 cm and 4 cm, respectively. If these two piles of gravel are submerged in the large pool, by how many centimeters will the water level in the large pool rise?
$$
1 \frac{17}{18}
$$
3.【Solution】When the gravel is submerged in water, the increase in the water level corresponds to the volume of the submerged gravel. Therefore, the volume of the gravel submerged in the pool is: $3 \times 3 \times 0.06 = 0.54\left(\right.$ cubic meters $\left.\right)$,

and the volume of the gravel submerged in the small pool is: $2 \times 2 \times 0.04 = 0.16\left(\right.$ cubic meters $\left.\right)$,

The total volume of these two piles of gravel is: $0.54 + 0.16 = 0.7\left(\right.$ cubic meters $\left.\right)$

When they are all submerged in the large pool, the increase in the water level in the large pool is also 0.7 cubic meters.

The base area of the large pool is: $6 \times 6 = 36\left(\right.$ square meters $\left.\right)$,

So the water level in the large pool rises by: $0.7 \div 36 = \frac{0.7}{36}$ (meters) $= \frac{70}{36}$ (centimeters) $= 1 \frac{17}{18}$ (centimeters). Answer: The water level in the large pool rises by $1 \frac{17}{18}$ centimeters.