To find the volume of the water in the container, we need to consider the volume occupied by the water as well as the volume displaced by the balls.

1. Compute the total volume of the container:
   The volume $V$ of a rectangular container is given by the product of its length $l$, width $w$, and height $h$.
   $V = l \times w \times h = 15 \, \text{cm} \times 9 \, \text{cm} \times 10 \, \text{cm} = 1350 \, \text{cm}^3$

2. Compute the volume displaced by the two balls:
   The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$. For a ball with a radius $r = 3 \text{cm}$:
   $V_{\text{ball}} = \frac{4}{3} \pi (3 \, \text{cm})^3 = \frac{4}{3} \pi (27 \, \text{cm}^3) = 36 \pi \, \text{cm}^3$ Hence, the volume of two such balls is:
   $2 \times 36 \pi \, \text{cm}^3 = 72 \pi \, \text{cm}^3$

3. Compute the current water volume with a depth of 3 cm:
   Since the water reaches a depth of 3 cm, its volume in the rectangular section is given by:
   $V_{\text{water}} = l \times w \times \text{depth} = 15 \, \text{cm} \times 9 \, \text{cm} \times 3 \, \text{cm} = 405 \, \text{cm}^3$

The volume of water initially in the container must have displaced the water level to account for the balls. Therefore, adding the water displaced by the balls to the present water volume will provide the total initial water volume:

4. Total initial volume of water in the container:
   $V_{\text{initial water}} = V_{\text{water}} + V_{\text{displaced by balls}} = 405 \, \text{cm}^3 + 72 \pi \, \text{cm}^3$

Note that $72 \pi \, \text{cm}^3$ equals approximately $226.08 \, \text{cm}^3$ (using $\pi \approx 3.14$), therefore:

• $V_{\text{initial water}} \approx 405 \, \text{cm}^3 + 226.08 \, \text{cm}^3 = 631.08 \, \text{cm}^3$

Thus, the initial volume of water in the container was approximately $631.08 \, \text{cm}^3$.