Answer
0.491 m
Work Step by Step
See figure 12-11. For a pipe open at both ends, the fundamental frequency is $f_1=\frac{v}{2 \mathcal{l}}$. The length is the distance from the mouthpiece (antinode) to the first open hole in the flute tube (another antinode).
$$\mathcal{l}=\frac{v}{2f_1}=\frac{343m/s}{2(349Hz)}=0.491m $$
