If your problem was a little bit more difficult (roland-bacher already provided an easy, precise and correct solution) and your $n$ is big, you can also approximate the binomial distribution by a Poisson distribution. Repeating the test $m$ times gives then the parameter $\lambda = \frac{m}{n}$ and your goal is that $m$ is big enough that $e^{-\lambda}\ge 1-p$. So $m \ge -n\ln(1-p)$. The approximation by the Poisson distribution is pretty good, for $n = 500$ and $p = 0.99$ it yields $m > 2302.585$ instead of the correct $m > 2300.28$ given by roland-bacher's formula.