Using the relation of Young's modulus,

 $\frac{T}{A}=Y\left(\frac{\Delta L}{L}\right)$
$\Rightarrow T=\left(\frac{Y\Delta L}{L}\times A\right)$

The linear mass density is $\mu =\left(\frac{m}{L}\right)$.

So,
$\frac{T}{\mu }=\frac{Y\Delta LA}{L\left(\frac{m}{L}\right)}=\frac{Y\left(\Delta L\right)\times LA}{L\left(m\right)}=\left(\frac{Y\Delta L}{L}\right)\times \left(\frac{1}{\rho }\right)$

Substituting the values,
$\frac{T}{\mu }=\frac{8\times {10}^{10}\times 3.2\times {10}^{-4}}{0.5}\times \left(\frac{1}{8\times {10}^3}\right)=6.4\times {10}^3$
$\Rightarrow \frac{T}{\mu }=64\times {10}^2$

The fundamental frequency is given by $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$.
$\Rightarrow \sqrt{\frac{T}{\mu }}=8\times 10=80 \mathrm{m} {\mathrm{s}}^{-1}$
Therefore,

$f=\left(\frac{80}{1}\right)=80 \mathrm{Hz}$