The time taken to do the $n^{th}$ round is $\dfrac{a(n)}{(75+50)}$ where $a(n)$ is the distance between the two trains at the beginning of the $n^{th}$ round.
Note that $$a(n+1) = a(n) - (50+50) \times \dfrac{a(n)}{125} = \dfrac{a(n)}{5} = \dfrac{a(1)}{5^n}.$$
The distance travelled by the fly in the $n^{th}$ round is $75 \times \dfrac{a(n)}{125} = \dfrac{3}{5} a(n)$.
Hence, the total distance travelled by the fly is $$\sum_{n=1}^{\infty} \dfrac{3}{5} a(n) = a(1) \dfrac{3}{5} \left( 1 + \dfrac15 + \dfrac1{5^2} + \cdots \right) = a(1) \dfrac{3}{5} \times \dfrac{5}{4} = \dfrac34 a(1)$$ which is nothing but $$\text{Speed of the fly} \times \underbrace{\dfrac{a(1)}{\text{Relative speed between the trains}}}_{\text{Time taken by the trains to collide}}$$
which makes sense.
If you want the total number of rounds the fly makes before the train collides, this amounts to the number of rounds till when $a(n) = 0$. However note that only as $n \rightarrow \infty$, $a(n) \rightarrow 0$. Hence, the fly will make $\infty$-rounds before the trains collide!
The time taken by the fly to make the first round is $$\dfrac{a(1)}{125}.$$ The time taken by the fly to make the $n^{th}$ round is $$\dfrac{a(n)}{125}.$$ Hence, the time taken till the $N^{th}$ round is $$\sum_{n=1}^{N} \dfrac{a(n)}{125} = \sum_{n=1}^{N} \dfrac{ \dfrac{a(1)}{5^{n-1}}}{125}$$
Hence, solve for $N$ in the equation $$\sum_{n=1}^{N} \dfrac{ \dfrac{a(1)}{5^{n-1}}}{125} = t$$
Setting $a(1) = 100$, we get that
$$\sum_{n=0}^{N-1} \left(\dfrac15 \right)^n = \dfrac{5t}{4}$$
$$\dfrac{\left(\dfrac15 \right)^N - 1}{\left(\dfrac15 \right) - 1} = \dfrac{5t}{4}$$
$$1 - \left(\dfrac15 \right)^N = t$$ $$\left(\dfrac15 \right)^N = 1-t$$ Hence the number of rounds, $N$, made by the fly as a function of time $t$ is $$N = \dfrac{\log(1-t)}{\log(1/5)}$$
