Coin flipping probability?
That's the right calculation: you want the smallest integer $n$ such that $0.5^n<0.005$. However, there is a potentially quicker way to work this out. Taking logs we get $n\log 0.5<\log 0.005$, so $n>\frac{\log 0.005}{\log 0.5}$ (the inequality reverses sign because $\log 0.5$ is negative), so the answer is $\frac{\log 0.005}{\log 0.5}$ rounded up to the next integer. $\frac{\log 0.005}{\log 0.5}\approx 7.64$, so this gives $n=8$.