A copper rod of length 1.0 m is clamped at its middle point....

The rod is clamped at its middle point, which means a node is formed at the center and antinodes are formed at the two free ends.

The distance from the clamped center to either free end is $\dfrac{L}{2}$. For standing waves, this distance must be an odd multiple of $\dfrac{\lambda}{4}$.

$\dfrac{L}{2} = (2n - 1)\dfrac{\lambda}{4}$

$\lambda = \dfrac{2L}{2n - 1}$

The corresponding frequencies are:

$f_n = \dfrac{v}{\lambda} = \dfrac{(2n - 1)v}{2L}$

Substituting $v = 3800 \text{ m s}^{-1}$ and $L = 1.0 \text{ m}$:

$f_n = \dfrac{(2n - 1) \times 3800}{2 \times 1.0} = 1900(2n - 1) \text{ Hz} = 1.9(2n - 1) \text{ kHz}$

The frequencies must lie between $20 \text{ Hz}$ and $20{,}000 \text{ Hz}$ ($20 \text{ kHz}$).

$1.9(2n - 1) \le 20$

$2n - 1 \le 10.52$

$2n \le 11.52 \Rightarrow n \le 5.76$

Since $n$ is an integer, $n = 1, 2, 3, 4, 5$.