We know that the wavelength is related to the speed of the wave by:
![v=\lambda f](https://img.qammunity.org/2023/formulas/physics/college/i838uwfyqotnoyo83z5m5nh9ah7vk1pyb2.png)
where f is the frequency.
The speed of sound on air at a given temperature is given by:
![v=331\sqrt[]{1+(T)/(273)}](https://img.qammunity.org/2023/formulas/physics/college/dx7fgpct0wcibxzocg0d9sz3ou1pe6ziax.png)
so in this case the speed is:
![v=331\sqrt[]{1+(22)/(273)}=344.08](https://img.qammunity.org/2023/formulas/physics/college/s6n8rodbkkxhuycgzhhu9mo5ldwgmmmfhh.png)
Plugging this and the frequency in the first expression above we have:
![\begin{gathered} 344.08=90*10^3\lambda \\ \lambda=(344.08)/(90*10^3) \\ \lambda=3.82*10^(-3) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/tttrmg31p6iqkh876add2mwh8ago2dwwih.png)
Therefore the wavelength is:
![3.82*10^(-3)\text{ m}](https://img.qammunity.org/2023/formulas/physics/college/whv6uc14zbi2sssjce7lw4k7ri0gg5vl34.png)