Answer
$36\;\rm cm$
Work Step by Step
The author told us to ignore the thin glass of the bowl, so we have here two indexes of refraction, $n_{water}$ and $n_{air}$.
We need to find out how far from the edge did the fish see the cat.
This means that $n_1=n_{air}$ and $n_2=n_{water}$.
According to the fish, this is a convex spherical refracting surface, so
$$\dfrac{n_1}{s}+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}$$
The cat's face is 20 cm from the bowl which means that, relative to the fish, $s=20$ cm.
Solving for $s′$,
$$ \dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}-\dfrac{n_1}{s}$$
$$ {s'}=n_2\left[\dfrac{n_2-n_1}{R}-\dfrac{n_1}{s}\right]^{-1}$$
Plugging the known;
$$ {s'}=(1.33)\left[\dfrac{1.33-1}{25}-\dfrac{1}{20}\right]^{-1}=\color{red}{\bf -36.1}\;\rm cm$$
So the fish sees the cat as a virtual image 36 cm outside away from the edge of the bowl.
