${\mathit{i}}_1={\mathit{i}}_2=3\mathrm{A}$
$\mathit{r}=2\mathrm{m}$
$\mathit{F}∕\mathit{l}=\frac{{\mathit{\mu }}_0}{2\mathit{\pi }}\times \frac{{\mathit{i}}_2{\mathit{i}}_2}{\mathit{r}}$
$\left[\frac{{\mathit{\mu }}_0}{2\mathit{\pi }}=2\times {10}^{-7}{\mathit{N}}_{\mathit{A}}^{-2}\right]$
Or, $\mathit{F}∕\mathit{l}=2\times {10}^{-7}\times \frac{3\times 3}{2}$
∴ $\mathit{F}∕\mathit{l}=9\times {10}^{-7}\mathrm{N}$
Force per unit length $=9.0\times {10}^{-7}\mathrm{N}∕\mathrm{m}$
Since, the currents are in same direction so there would be an attractive force between them.