The common ratio of the geometric sequence is $r=\frac{15}{\frac{3}{4}}=20$.
So, the $n$th term of the sequence is $\left(\frac{3}{4}\right)(20^{n-1})$.
We want to find the smallest $n$ such that $\left(\frac{3}{4}\right)(20^{n-1})$ is divisible by one million.
In other words, we want to find the smallest $n$ such that $20^{n-1}$ is divisible by $\frac{4}{3}\times 10^6$.
Since $\frac{4}{3}\times 10^6$ is a multiple of $2^6$ and $5^6$, we need $20^{n-1}$ to be divisible by $2^6$ and $5^6$.
The smallest $n$ that satisfies both conditions is $n=\boxed{7}$.
The answer is: 7
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Let the first term of a geometric sequence be $\frac{3}{4}$, and let the second term be $15$. What is the smallest $n$ for which the $n$th term of the sequence is divisible by one million?

### Answer:
