A cylindrical jar of radius 6 cm holds water to a depth of 9 cm…

Step 1. Is it covered? The diameter of the ball is $6\ \text{cm}$. The new depth will be more than $9\ \text{cm}$, so the ball is covered.

Step 2. The volume of the ball. $\dfrac{4}{3}\pi \times 27\ \text{cm}^3 = 36\pi\ \text{cm}^3$.

Step 3. The surface of the water. $\pi \times 36\ \text{cm}^2 = 36\pi\ \text{cm}^2$.

Step 4. The rise. $\dfrac{36\pi\ \text{cm}^3}{36\pi\ \text{cm}^2} = 1\ \text{cm}$.

Step 5. The new depth. $9\ \text{cm} + 1\ \text{cm} = 10\ \text{cm}$.