If you are expected to do this kind of problem then you should know that, given a normal distribution with mean $\mu$ and standard deviation $\sigma$, variable x corresponds to a "standard normal distribution" $z= \frac{x- \mu}{\sigma}$. Here, you need to look up, in a table of the standard normal distribution, $z_0$ corresponding to $P(z\ge z_0)= 0.0218$ and $z_1$ such that $P(z\ge z_1)= 0.9345$. Then, using those values for $z_0$ and $z_1$, solve the two equations $z_0= \frac{59.1- \mu}{\sigma}$ and $z_1= \frac{29.2- \mu}{\sigma}$ for $\mu$ and $\sigma$.