I did not test every number up to 750,000, but seeing the patterns, I suspect this is the answer.

I used a recursive program for Fibonacci which saves to a dictionary.

If I start with too large a number, the  program chokes on recursion depth.  But if I start small enough and increase gradually, it was able to go high enough.

At first, using smaller parameters, I found the patterns below, and eventually found that:

Patterns: 

Every 15th Fib is a multiple of 10

Every 150th Fib is a multiple of 100

Every 750th Fib is a multiple of 1000

Fib(7500) ends in 4 zeros

Fib(75000) ends in 5 zeros

Fib(150000) also ends in 5 zeros

Fib(750000) ends in 6 zeros

This number is 15674 digits long

The program output was quite long, here is the start and end.

The second number of the pair is the number of zeros at the end of that Fibonacci number.

[[750, 3], [1500, 3], [2250, 3], [3000, 3], [3750, 3], [4500, 3], [5250, 3], [6000, 3], [6750, 3], [7500, 4], [8250, 3], [9000, 3], [9750, 3], [10500, 3], [11250, 3], [12000, 3], [12750, 3], [13500, 3], [14250, 3], [15000, 4], [15750, 3],  ... ... ,

...  ...

[741750, 3], [742500, 4], [743250, 3], [744000, 3], [744750, 3], [745500, 3], [746250, 3], [747000, 3], [747750, 3], [748500, 3], [749250, 3], [750000, 6]]

----------------------

def endzeros(n):

    """ how many zeros at the end of n """

    s = str(n)[::-1]

    count = 0

    for i,v in enumerate(s):

        if v == '0':

            count += 1

        else:

            return count

    return count

d = {0:0, 1:1, 2:1}

def fib(n):

    """ Efficient Fibonacci using dictionary,

    but not requiring dictionary to be already defined """

    global d

    if 'd' not in globals(): 

        d = {1:1, 2:1}

    if n in d:

        return d[n]

    else:

        ans = fib(n-1) + fib(n-2)

        d[n] = ans

        return ans

listanswers = []

for i in range(750,750001,750):

    listanswers.append([i, endzeros(fib(i))])

print(listanswers)