According to the right-hand screw rule, the magnetic field will be into the plane across the loop.
Force on length AD:
F = Bil
$\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi\text{x}}$ (Towards Right)
Force on length AD:
F = Bil
$\text{F}_2=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi(\text{x}+\text{a})}$ (Towards Left)
Force on AB and CD will be equal and opposite.
Hence, they'll cancel out.
Force on the loop:
$\text{F}_\text{Net}=\text{F}_1-\text{F}_2$
$=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi}\Big[\frac{1}{\text{x}}-\frac{1}{(\text{x}+\text{a})}\Big]$
$\text{F}_\text{Net}=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi}\Big[\frac{\text{x}+\text{a}-\text{x}}{(\text{x}+\text{a})\text{x}}\Big]=\frac{\mu_0\text{I}_1\text{I}_2\text{a}^2}{2\pi(\text{x}+\text{a})\text{x}}$
$\text{F}_\text{Net}=\frac{\mu_0\text{I}_1\text{I}_2\text{a}^2}{2\pi\text{x}(\text{x}+\text{a})}$ (Towards left)
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