Answer
$\approx13\;\rm Hz$
Work Step by Step
We know that the fundamental frequency in a wire is given by
$$f_{(m=1)}=\dfrac{v}{2L}$$
$$f_{1}=\dfrac{1}{2L}\sqrt{\dfrac{T_s}{\mu}}\tag 1$$
where $L$ is the length of the wire, $T_s$ is the tension in the wire, and $\mu$ is the linear density of the wire.
Now we need to find the length of the wire $L$, so we need to use the Pythagorean theorem.
$$L=\sqrt{2^2+2^2}=\bf2\sqrt{2}\;\rm m\tag 2$$
Noting that this angle is an Isosceles triangle since the two non-right angles are 45$^\circ$.
Now we need to find the tension in the wire and since the system is in static equilibrium,
$$\sum \tau=0$$
We chose the left side of the horizontal rod as the point at which we tried to find the torque around it. We chose counterclockwise to be our positive torque direction.
$$\sum \tau=(-8\times 9.8\times 2) +(T\sin45^\circ \times 2)+(-4\times 9.8\times 1.0)=0$$
Hence,
$$T=\bf 138.6\;\rm N\tag 3$$
Now we need to find $\mu$ of the wire,
$$\mu=\dfrac{m}{L}=\dfrac{75\times 10^{-3}}{2\sqrt{2}}=\bf 0.0265\;\rm kg/m\tag 4$$
Plugging (2), (3), and (4) into (1);
$$f_{1}=\dfrac{1}{2(2\sqrt{2})}\sqrt{\dfrac{138.6}{0.0265}} =\color{red}{\bf12.8}\;\rm Hz$$
