\((2^5)\times(6^2)\times(7^3)\times n\) is the given number.
If both \(5^2,3^3\) are factors, then they must be present in the number.
Leaving rest of the prime factors and splitting \(6^2\) into \(3^2\times2^3\).
The number is lacking \(5^2\) and a3, so that \(5^2\) and \(3^3\) is a factor.
Hence the smallest number is \(5^2\times3=75\)
The correct option is (D): 75