Answer
$\frac{9}{4}\mu_0$
Work Step by Step
As we see, in the given figure, the two wires are having the same length, the same frequency, and the same tension since both are connected by the same spring.
So,
$$f_{\rm left}=f_{\rm right}\tag 1$$
where
$$(f_{m=2})_{\rm left}=\dfrac{2v}{2L}=\dfrac{1}{L}\sqrt{\dfrac{T_s}{\mu_0}}$$
and
$$(f_{m=3})_{\rm right}=\dfrac{3v}{2L}=\dfrac{3}{2L}\sqrt{\dfrac{T_s}{\mu}}$$
Thus,
$$ \dfrac{1}{ \color{red}{\bf\not} L}\sqrt{\dfrac{ \color{red}{\bf\not} T_s}{\mu_0}} =\dfrac{3}{2 \color{red}{\bf\not} L}\sqrt{\dfrac{ \color{red}{\bf\not} T_s}{\mu}}$$
Squaring both sides;
$$\dfrac{1}{\mu_0}=\dfrac{9}{4\mu}$$
Therefore,
$$\boxed{\mu=\frac{9}{4}\mu_0}$$
