Force acting between the wires will be attractive as current flowing in both wires is in the same direction.

Now to keep the wire hanging in air, force acting per unit length of the wire due to magnetic field should balance gravitational force. Therefore,

$$\begin{gathered} F=\lambda g \\ \Rightarrow \frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\lambda g \\ \Rightarrow I_2=\frac{2\pi d\lambda g}{{\mu }_0{I}_1} \\ =\frac{2\pi \left(4\times {10}^{-2}\right)\left(10\times {10}^{-3}\right)\times 10}{4\pi \times {10}^{-7}\times 160} \\ =125 \mathrm{A} \end{gathered}$$