A container is filled with water (μ = 4/3 ) up to a height of 24...

Let the apparent position of the object $O$ as seen by the mirror be $O'$.

Apparent depth of object $O$ from the water surface is $d_{ap} = \dfrac{d_{real}}{\mu} = \dfrac{24}{4/3} = 18 \text{ cm}$.

The distance of $O'$ from the concave mirror is $u = - (10 + 18) = -28 \text{ cm}$.

Let the apparent position of the image $I$ as formed by the mirror, seen from the water surface, be $I'$.

The image $I$ is formed $20 \text{ cm}$ below the water level.

Apparent depth of image $I'$ from the water surface is $d'_{ap} = \dfrac{20}{4/3} = 15 \text{ cm}$.

The distance of $I'$ from the concave mirror is $v = - (10 + 15) = -25 \text{ cm}$.

Using the mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{-25} + \dfrac{1}{-28} = \dfrac{1}{f}$

$\dfrac{1}{f} = -\left(\dfrac{28 + 25}{25 \times 28}\right) = -\dfrac{53}{700}$

$f = -\dfrac{700}{53} \approx -13.2 \text{ cm}$

Answer: $-13.2\ cm$