Answer
$ \rm 85.2 \;cm,\; 55.6 \;cm, 26.1 \;cm$
Work Step by Step
It is an open-closed tube model where the frequency is then given by
$$f_m=\dfrac{mv}{4L}\tag{$m=1,3,5,..$}$$
where $L$ here is the empty part of the tube when the standing waves occur.
Hence,
$$L=\dfrac{mv}{4f_m}=\dfrac{343m}{4(580)}$$
where $v$ and $f_m$, in this case, are constants since the sound of the wave is constant and the frequency of the tuning fork is also constant.
Noting that the height of water at which there are standing waves is then given by
$$h=1-L=1-\dfrac{343m}{4(580)}$$
It is better to put the data in a table as seen below by plugging the values of $m$;
\begin{array}{|c|c|c|}
\hline
m&h\;\rm (cm) \\
\hline
1& 85.2\\
\hline
3& 55.6\\
\hline
5& 26.1 \\
\hline
7& -3.5\;{\rm \bf Dismissed} \\
\hline
\end{array}
So we only have 3 possible heights at which there are standing waves,
$$h=\rm \color{red}{\bf 85.2}\;cm,\; \color{red}{\bf55.6}\;cm, \color{red}{\bf26.1}\;cm$$
