For a pipe open at both ends, the resonant frequencies are given by:
\[
f_n = n \frac{v}{2L}
\]
where \(n\) is a positive integer (harmonic number), \(v\) is the velocity of sound, and \(L\) is the length of the pipe.
Given:
\[
L = 0.3 \, {m}, \quad v = 330 \, {m/s}, \quad f = 1.65 \, {kHz} = 1650 \, {Hz}
\]
Solving for \(n\):
\[
1650 = n \frac{330}{2 \times 0.3} \quad \Rightarrow \quad 1650 = n \times 550 \quad \Rightarrow \quad n = \frac{1650}{550} = 3
\]
Thus, the third harmonic mode resonates with a 1.65 kHz source.