Given data:
* The frequency of the tuning fork is,

* The wavelength of the fork at 20 degree is,

Solution:
The velocity of the tuning fork wave at 20 degree is,

The velocity of the tuning fork wave at 13 degree is,
![v=v_(\circ)\sqrt[]{(T)/(T_(\circ))}](https://img.qammunity.org/2023/formulas/physics/college/j5b7u77etbgm6fus94o1r1ee6ucpra24zh.png)
where,

and,

Thus, the velocity of the sound wave at 13 degree Celsius is,
![\begin{gathered} v=343.2*\sqrt[]{(286.15)/(293.15)} \\ v=339.08\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/6urfn9p17jal77mj1cnh50erupezl867j5.png)
The wavelength of tthe sound wave at 13 degree celsius is,

Substituting the known values,

Thus, the wavelength of the sound wave at 13 degree celsius is 0.77 m.
Hence, option 1 is the correct answer.