Answer
The distance to the cat appears to be more than the actual distance.
Work Step by Step
We can use this formula:
$d' = (\frac{n_1}{n_2})~d$
where $d$ is the actual distance and $d'$ is the apparent distance.
In this case, since the fish is in the water and the cat is outside the water, $n_1 = 1.33$ and $n_2 = 1.00$
Therefore: $~~d' \gt d$
The distance to the cat appears to be more than the actual distance.
