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+### Statement
+
+$3.9.15.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$3.9.15.$ Determine the first resonant frequency of air oscillations between two parallel buildings located at a distance $L = 20 \text{ m}$ from each other. The height of the buildings is noticeably greater than this distance. The speed of sound in air is $c = 330 \text{ m/s}$.
+
+### Solution
+
+The first resonant frequency (fundamental tone) corresponds to the formation of a standing sound wave, where displacement nodes are formed at the boundaries (the rigid walls of the buildings). This means that exactly half of the wavelength $\lambda$ fits between the walls:
+$$L = \frac{\lambda}{2}$$
+
+From here, we can find the wavelength:
+$$\lambda = 2L = 2 \cdot 20 \text{ m} = 40 \text{ m}$$
+
+Considering the relationship between the wavelength, the speed of sound $c$, and the frequency $\nu$ ($c = \lambda \nu$), we can express the resonant frequency:
+$$\nu = \frac{c}{\lambda} = \frac{330 \text{ m/s}}{40 \text{ m}} = 8.25 \text{ Hz}$$
+
+#### Answer
+$$\nu = 8.25 \text{ Hz}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]