Answer
See the detailed answer below.
Work Step by Step
Recall that for parallel current carrying wires, parallel currents attract and opposite currents repulsive.
So, the current from the two lower wires will push the upper wire upward.
The net force exerted on the upper wire is given by
$$\sum F_{y,\rm on1}=F_{\rm 2on1}\cos\theta+F_{3on1} \cos\theta-mg=ma_y=m(0)$$
$$F_{\rm 2on1}\cos\theta+F_{3on1} \cos\theta=mg\tag 1$$
Recalling that the force exerted by one wire on the other is given by
$$F=\dfrac{\mu_0 L_1I_1I_2}{2\pi d}$$
Plug into (1),
$$\dfrac{\mu_0 L_1I_1I_2}{2\pi d_{2to1}}\cos\theta+\dfrac{\mu_0 L_1I_1I_3}{2\pi d_{3to1}} \cos\theta=mg $$
where we need the 3 currents to be equal in magnitude, so $I_1=I_2=I_3$, and we need the 3 wires to form an equilateral triangle, so $d_{2to1}=d_{3to1}=d_{2to3}=d$
$$\dfrac{\mu_0 L_1I^2}{2\pi d }\cos\theta+\dfrac{\mu_0 L_1I^2}{2\pi d} \cos\theta=mg $$
Hence,
$$I^2\left[ \dfrac{2\mu_0 L_1 \cos\theta}{2\pi d } \right]=mg $$
$$I^2 =\dfrac{mg}{\dfrac{ \mu_0 L_1 \cos\theta}{ \pi d } } $$
$$I =\sqrt{\dfrac{\pi mgd}{ \mu_0 L_1 \cos\theta } }$$
We know that $m=\mu L$, where $\mu$ here is the linear mass density of the wire.
$$I =\sqrt{\dfrac{\pi \mu L_1gd}{ \mu_0 L_1 \cos\theta } }$$
$$I =\sqrt{\dfrac{\pi \mu gd}{ \mu_0 \cos\theta } }$$
From the geometry of the figure below, $\theta=30^\circ$, plug the known;
$$I =\sqrt{\dfrac{\pi (50\times 10^{-3})(9.8)(0.04)}{ (4\pi \times 10^{-7})\cos30^\circ} }$$
$$I=\color{red}{\bf 238}\;\rm A$$
