Answer
$93\;\rm m$
Work Step by Step
This is an open-closed tube problem. We need to find the length of the stunned where the frequencies of the standing waves inside it are given by
$$f_m=\dfrac{mv}{4L}\tag{$m=1,3,5,...$}$$
And since we are dealing with only the odd numbers of $m$ since it is an open-closed tube model, the two next frequencies are at $m$, and $m+2$.
So,
$$f_m=\dfrac{mv}{4L}$$
$$f_{(m+2)}=\dfrac{(m+2)v}{4L}$$
Hence,
$$f_{(m+2)}-f_m=\dfrac{(m+2)v}{4L}-\dfrac{mv}{4L}=\dfrac{2v}{4L}=\dfrac{v}{2L}$$
$$f_{(m+2)}-f_m=\dfrac{v}{2L}$$
Thus,
$$L=\dfrac{v}{2[f_{(m+2)}-f_m]}$$
Plugging the two given frequencies and the speed of sound in the tunnel;
$$L=\dfrac{335}{2[6.3-4.5]}=\color{red}{\bf 93}\;\rm m$$
