As shown in the figure, a square loop PQRS carrying a current of...

The magnetic force on a straight wire of length $L$ carrying current $I_2$ placed at a distance $r$ from a long straight wire carrying current $I_1$ is given by:

$F = \dfrac{\mu_0 I_1 I_2 L}{2\pi r}$

The forces on the horizontal segments $PQ$ and $RS$ are equal and opposite, and their net force is zero.

For the side $SP$ of the square loop:
The current in $SP$ is downwards, which is anti-parallel to the current in the long wire. Therefore, the force on $SP$ is repulsive (towards the right).
Distance $r_1 = 1\text{ cm} = 0.01\text{ m}$
Length $L = 2\text{ cm} = 0.02\text{ m}$
$F_{SP} = \dfrac{4\pi \times 10^{-7} \times 10 \times 6 \times 0.02}{2\pi \times 0.01} = 2.4 \times 10^{-5}\text{ N}$ (towards right)

For the side $RQ$ of the square loop:
The current in $RQ$ is upwards, which is parallel to the current in the long wire. Therefore, the force on $RQ$ is attractive (towards the left).
Distance $r_2 = 1\text{ cm} + 2\text{ cm} = 3\text{ cm} = 0.03\text{ m}$
$F_{RQ} = \dfrac{4\pi \times 10^{-7} \times 10 \times 6 \times 0.02}{2\pi \times 0.03} = 0.8 \times 10^{-5}\text{ N}$ (towards left)

The net magnetic force on the loop is:
$F_{net} = F_{SP} - F_{RQ} = 2.4 \times 10^{-5} - 0.8 \times 10^{-5} = 1.6 \times 10^{-5}\text{ N}$

Since $F_{SP} > F_{RQ}$, the net force is directed towards the right.