Given the radius of the spherical ball r = 1.5 cm:

The volume of a sphere is given by the formula $V_{sphere} = \frac{4}{3} \pi r^3$.
Given the radius of the spherical ball $r = 1.5\, \text{cm}$:

$$V_{sphere} = \frac{4}{3} \pi (1.5\, \text{cm})^3$$ $$V_{sphere} = \frac{4}{3} \pi (3.375\, \text{cm}^3)$$ $$V_{sphere} = 4.5 \pi\, \text{cm}^3$$

When the spherical ball is immersed, it displaces a volume of water equal to its own volume. This displaced water causes the water level in the cylindrical tin to rise.
Let $h$ be the rise in the water level. The volume of this rise is a cylinder with the radius of the tin and height $h$.
The volume of a cylinder is given by the formula $V_{cylinder} = \pi R^2 h$.
Given the radius of the cylindrical tin $R = 1.8\, \text{cm}$:

$$V_{rise} = \pi (1.8\, \text{cm})^2 h$$ $$V_{rise} = \pi (3.24\, \text{cm}^2) h$$

The volume of the sphere is equal to the volume of the water rise:

$$V_{sphere} = V_{rise}$$ $$4.5 \pi\, \text{cm}^3 = 3.24 \pi\, \text{cm}^2 h$$

Divide both sides by $\pi$:

$$4.5\, \text{cm}^3 = 3.24\, \text{cm}^2 h$$

Solve for $h$:

$$h = \frac{4.5\, \text{cm}^3}{3.24\, \text{cm}^2}$$ $$h \approx 1.3888... \, \text{cm}$$

$$h \approx 1.4\, \text{cm}$$

The rise in the water level in the tin is $\text{1.4 cm}$.