First, because 

$$\lim_{x\to 0}\frac{\sin^nx}{\cos^2x(1-\cos{x})^3}=\lim_{x\to 0}\frac{\sin^nx}{(1-\cos{x})^3}$$

therefore, we can safely remove $\cos^2x$ in the denominator from consideration. Now, by Taylor expansion, we have

$$\sin{x} = x + O(x^3)\qquad\text{and}\qquad (1-\cos{x}) = \frac{x^2}{2!} + O(x^4)$$

It follows that

$$\lim_{x\to 0}\frac{\sin^nx}{(1-\cos{x})^3}=\lim_{x\to 0}\frac{x^n}{(0.5x^2)^3}$$

For the last limit to exist, the exponent of $x$ in the numerator must be at lease equal to that in the denominator. Hence, the answer is $\boxed{6}$.