A container holds water up to a height of 20 cm and a point source...

Let the angle of incidence of the light ray at the edge of the rubber ring be $i$.

From the geometry of the setup, we have:
$\tan i = \dfrac{r}{h} = \dfrac{15 \text{ cm}}{20 \text{ cm}} = \dfrac{3}{4}$

This gives $\sin i = \dfrac{3}{5}$.

Applying Snell's law at the water-air interface:
$\mu_{w} \sin i = \mu_{a} \sin r'$

$\dfrac{4}{3} \times \dfrac{3}{5} = 1 \times \sin r'$

$\sin r' = \dfrac{4}{5} \Rightarrow \tan r' = \dfrac{4}{3}$

The light ray emerges into the air at an angle $r'$ and travels to the ceiling, which is at a height $H = 2.0 \text{ m}$ above the water surface. The additional horizontal distance traveled by the ray in the air is:
$x = H \tan r' = 2.0 \times \dfrac{4}{3} = \dfrac{8}{3} \text{ m} \approx 2.667 \text{ m}$

The total radius of the shadow formed on the ceiling is the sum of the radius of the ring and the horizontal distance traveled in the air:
$R = r + x = 0.15 \text{ m} + 2.667 \text{ m} = 2.817 \text{ m}$

Rounding off to two decimal places, we get $R \approx 2.82 \text{ m}$.