Answer
8.6 m to $8.6\times 10^{-3}m$.
Work Step by Step
See figure 12-11. For a pipe open at both ends, the fundamental frequency is $f_1=\frac{v}{2 \mathcal{l}}$. Solve for the length.
$$\mathcal{l}=\frac{v}{2f_1}$$
Find the length of the pipe needed for the highest and lowest frequencies.
$$\mathcal{l}_{20Hz}=\frac{343m/s}{2(20Hz)}=8.6m$$
$$\mathcal{l}_{20kHz}=\frac{343m/s}{2(20000Hz)}=8.6\times 10^{-3}m$$