To solve this problem, we need to understand the harmonic frequencies of a tube closed at one end.$\\$For a tube closed at one end, the harmonics are given by:$\\ f_n = n \cdot f_1 \quad$where $n$ is an odd integer (1, 3, 5, ...)$\\$Given that the two nearest harmonics are 220 Hz and 260 Hz, we can write:$\\ f_n = 220 \,$Hz$\\ f_{n+2} = 260 \,$Hz$\\$The difference between consecutive odd harmonics is twice the fundamental frequency:$\\ f_{n+2} - f_n = 2 \cdot f_1 \\$Substitute the given values:$\\ 260 - 220 = 2 \cdot f_1 \\$Simplify to find the fundamental frequency:$\\ 40 = 2 \cdot f_1 \\ f_1 = \frac{40}{2} \\ f_1 = 20 \,$Hz$\\$Therefore, the fundamental frequency of the system is $20 \,$Hz$. \\$This corresponds to Option 1.