Answer
$7.1\;\rm A$
Work Step by Step
First of all, we need to draw the force diagram of any wire of both wires since both wires are having the same current.
As you see below, they are 3 forces acting on the right wire.
The net force on any wire is zero since the whole system is at rest.
Thus,
The net force exerted on the wire in the $x$-direction is zero.
$$\sum F_x=F_B-T\sin3^\circ=ma_x=m(0)=0$$
$$F_B-T\sin3^\circ=0$$
$$F_B=T\sin3^\circ\tag 1$$
we know that the magnetic force between two wires is given by
$$F_B=B_{2on1}l_1I_1=\dfrac{\mu I_2}{2\pi r}I_1l_1$$
$$F_B=\dfrac{\mu I^2l}{2\pi S}\tag 2$$
Note that the two wires are having the same current and the have the same length. Recall that $(r=S)$ is the distance perpendicular between the two wires.
And the net force exerted on the wire in the $y$-direction is also zero.
$$\sum F_y=T\cos3^\circ-mg=ma_y=m(0)=0$$
Thus,
$$T\cos3^\circ=mg $$
Solving for $T$;
$$T=\dfrac{mg}{\cos3^\circ} $$
Plugging into (1);
$$F_B=\dfrac{mg}{\cos3^\circ}\;\sin3^\circ=mg\tan3^\circ$$
Plugging into (2);
$$\dfrac{\mu I^2l}{2\pi r}=mg\tan3^\circ$$
Solving for $I$;
$$ I^2 =\dfrac{2\pi S mg\tan3^\circ}{\mu l}$$
$$ I =\sqrt{\dfrac{2\pi S mg\tan3^\circ}{\mu l}}\tag 3$$
Now we need to find the mass of the wire which is given by the density law.
$$\rho=\dfrac{m}{V}$$
Thus,
$$m=\rho V=\rho A l$$
whereas $A$ is the cross-sectional area of the wire and $l$ is its length.
So, $A=\pi r^2$
$$m =\pi r^2 \rho l$$
Plugging into (3);
$$ I =\sqrt{\dfrac{2\pi S [\pi r^2 \rho l]g\tan3^\circ}{\mu l}}=\sqrt{\dfrac{2\pi^2 r^2 \rho S g\tan3^\circ}{\mu }}\tag 4$$
The distance between the wire is given from the geometry of the given figure,
$$\sin3^\circ=\dfrac{\frac{1}{2}S}{0.5}$$
Hence,
$$S=\sin3^\circ$$
Plugging into (4);
$$ I =\sqrt{\dfrac{2\pi^2 r^2 \rho g\sin3^\circ\tan3^\circ}{\mu }} $$
Plugging the known;
$$ I =\sqrt{\dfrac{2\pi^2 \left(\dfrac{0.42\times10^{-3}}{2}\right)^2 \cdot 2700\cdot 9.8 \cdot\sin3^\circ \tan3^\circ}{4\pi\times 10^{-7} }}$$
$$I=\color{red}{\bf 7.1}\;\rm A$$
