Answer:
13 years.
Explanation:
I don't understand the original equation you entered, so I'm going to use another one.

Where A = initial population
A₀ = current population
e = Euler's constant
k = rate of decrease
t = time in years
Let's start off with what we know.
Today there are 800 fish left:
A₀ = 800
There were 1800 fish 5 years ago:
t = -5
A = 1800
The equation now looks like

First we find the rate of decrease.

⇒

⇒

⇒
=k
⇒ k = -0.1621
The question asks when it will drop below 100, so make it equal to 100.
⇒

⇒

⇒

⇒
= t = 12.821
To the nearest year this is 13.