Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As we see in the sketch below, we treated the light as a ray model.
The rays inside the water from the fire are the refracted rays that penetrate the water-air surface at angles from 0$^\circ$ to 90$^\circ$ but of course, the refracted angles of the rays are from 0$^\circ$ to $\theta_c$ which is the critical angle between the water-air boundary.
So the fish just below the surface has to be nearer the fire-edge so it can see the fire. And the deeper it divs, the wider the range it can see the fire.
$$\color{blue}{\bf [b]}$$
Now we need to find the shallowest depth that the fish can see the fire. So we have to find the critical angle first which is given by Snell's law
$$n_{air}\sin90^\circ=n_{water}\sin\theta_c$$
So,
$$\theta_c=\sin^{-1}\left[ \dfrac{n_{air}}{n_{water}} \right]$$
$$\theta_c=\sin^{-1}\left[ \dfrac{1}{1.33} \right]=\bf 48.75^\circ$$
Hence, the smallest depth $y$ is given by
$$\tan\theta_c=\dfrac{20}{y}$$
Hence,
$$y=\dfrac{20}{\tan\theta_c}=\dfrac{20}{\tan 48.75^\circ}$$
$$y=\color{red}{\bf 17.5}\;\rm m$$
The fish must dive to 17.5 m so it can see the fire.
