Answer
$f_1\approx 19Hz$
$f_2=38Hz$
Work Step by Step
A tube closed at both ends will support standing waves with nodes at each end, so mathematically it resembles a string that is fixed at both ends. Therefore, the wavelength corresponding to the fundamental frequency is twice the length of the tube.
$$\lambda_1=2\mathcal{l}=2(9.0m)=18m$$
$$f_1=\frac{v}{\lambda_1}=\frac{343m/s}{18m}\approx 19Hz$$
$$f_2=2f_1=38Hz$$
