It will be 2.
because here the prime factorization can also be written as,
(2^2*5^2)*2*3^2*7
=(10^2)*2*3^2*7
=100 *2*3^2*7
Thus n will end with 2 consecutive zeroes.
UPvote if that helps:)

Prime Factorization of n

The prime factorization of a natural number n is the expression of that number as the product of its prime factors. In this case, the prime factorization of n is given as:

n = 2^3 * 3^2 * 5^2 * 7

Finding the Number of Consecutive Zeros

To determine the number of consecutive zeros in n, we need to find the highest power of 10 that divides n. Since 10 is a product of 2 and 5, we need to count the minimum number of times both 2 and 5 appear as factors in the prime factorization.

Counting Factors of 2
To count the factors of 2 in n, consider the exponent of 2 in the prime factorization. In this case, the exponent of 2 is 3. This means that 2^3 is the highest power of 2 that divides n.

Counting Factors of 5
To count the factors of 5 in n, consider the exponent of 5 in the prime factorization. In this case, the exponent of 5 is 2. This means that 5^2 is the highest power of 5 that divides n.

Counting Factors of 10
Since 10 is a product of 2 and 5, we need to find the minimum exponent between the factors of 2 and 5. In this case, the minimum exponent is 2, which corresponds to the factors of 5. Therefore, 10^2 is the highest power of 10 that divides n.

Number of Consecutive Zeros
The number of consecutive zeros in n is equal to the exponent of the minimum factor between 2 and 5, which is 2 in this case. Therefore, there are 2 consecutive zeros in n.

Conclusion
To find the number of consecutive zeros in a natural number n, we need to determine the minimum exponent between the factors of 2 and 5 in the prime factorization of n. In this case, the prime factorization of n is given as 2^3 * 3^2 * 5^2 * 7, and the minimum exponent between the factors of 2 and 5 is 2. Therefore, there are 2 consecutive zeros in n.