$\mathrm{v}=5.85 \times 10^{3} \mathrm{m} / \mathrm{sec}$

since rod is clamped at middle fundamental wave shape is as follow

$\frac{\lambda}{2}=\mathrm{L} \Rightarrow \lambda=2 \mathrm{L}$

$\lambda=1.2 \mathrm{m}(\because \mathrm{L}=60 \mathrm{cm}=0.6 \mathrm{m}(\mathrm{given})$

Using $v=f \lambda$

$\Rightarrow \quad f=\frac{v}{\lambda}=\frac{5.85 \times 10^{3}}{1.2}$

$=4.88 \times 10^{3} \mathrm{Hz}=5 \mathrm{KHz}$