Answer
$100.2\;\rm A$
Work Step by Step
We know that the net force exerted on wire M is zero since it is stationary.
We also know that the forces exerted on wire M by the two wires N and P are repulsive forces, as you see in the figures below.
From the previous problem, we found that the two horizontal components of the two forces exerted by wire N and wire B are having the same magnitude and opposite direction, so they cancel each other.
We also can see that the two vertical components of the same forces are both upward.
Thus,
$$\sum F_y=F_{\rm by\;N}\cos30^\circ+F_{\rm by\;P}\cos30^\circ-mg=0$$
Hence,
$$ F_{\rm by\;N}\cos30^\circ+F_{\rm by\;P}\cos30^\circ=mg $$
$$ B_NI_ML_M\cos30^\circ+B_PI_ML_M\cos30^\circ=\rho_{Cu} Vg $$
$$ \dfrac{\mu_0 I_NI_ML_M\cos30^\circ}{2\pi d}+\dfrac{\mu_0 I_PI_ML_M\cos30^\circ}{2\pi d}=\rho_{Cu} AL_Mg $$
$$ \color{red} {L_M}I_M\left[ \dfrac{\mu_0 I_N \cos30^\circ}{2\pi d}+\dfrac{\mu_0 I_P \cos30^\circ}{2\pi d}\right]=\rho_{Cu} A\color{red} {L_M}g $$
$$ I_M\left[ \dfrac{\mu_0 I_N \cos30^\circ}{2\pi d}+\dfrac{\mu_0 I_P \cos30^\circ}{2\pi d}\right]=\rho_{Cu} (\pi r^2) g $$
Recall that $I_N=I_P=I$
$$ I_M=\dfrac {\pi g\rho_{Cu} r^2 }{ \dfrac{2\mu_0 I \cos30^\circ}{2\pi d} }=\dfrac{2\pi^2 g\rho_{Cu}d r^2}{2\mu_0 I \cos30^\circ} $$
Plugging the known;
$$ I_M= \dfrac{2\pi^2 \cdot 9.8\cdot 8900\cdot 3.8\times10^{-2} \cdot\left(\frac{1}{2}\times10^{-3}\right)^2}{2\cdot 4\pi\cdot 10^{-7}\cdot 75 \cdot \cos30^\circ} $$
$$ I_M=\color{red}{\bf 100.2}\;\rm A$$
