Answer
See the detailed answer below.
Work Step by Step
First, we need to define the direction of the electric field created by each wire, so we need to use the right-hand rule, as seen in the figure below.
The force exerted on wire 1 is given by
$$\sum F_{\rm on1}=F_{\rm 2on1}+F_{3on1}$$
Recalling that for parallel wires and currents, parallel currents attract and opposite currents repulsive.
So the force exerted on wire 1 by wire 2 is upward, while the force exerted on wire 1 by wire 3 is downward.
So that,
$$\sum F_{\rm on1}=F_{\rm 2on1}-F_{3on1}$$
Recalling that the force exerted by one wire on the other is given by
$$F=\dfrac{\mu_0 L_1I_1I_2}{2\pi d}$$
$$\sum F_{\rm on1}=\dfrac{\mu_0 L_1I_1I_2}{2\pi d_{1\rightarrow 2}}-\dfrac{\mu_0 L_1I_1I_3}{2\pi d_{1\rightarrow 3}}$$
Plug the known;
$$\sum F_{\rm on1}=\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.02) }-\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.04) }$$
$$\sum F_{\rm on1}=(\color{red}{\bf 2.5\times 10^{-4}}\;{\rm N})\;\hat j$$
By the same approach,
The force exerted on wire 2 is given by
$$\sum F_{\rm on2}=F_{\rm 1on2}+F_{3on2}$$
So the force exerted on wire 2 by wire 1 is downward, while the force exerted on wire 2 by wire 3 is upward.
So that,
$$\sum F_{\rm on2}=F_{3on2}-F_{\rm 1on2}$$
$$\sum F_{\rm on2}=\dfrac{\mu_0 L_2I_2I_3}{2\pi d_{2\rightarrow 3}}-\dfrac{\mu_0 L_2I_1I_2}{2\pi d_{1\rightarrow 2}} $$
Plug the known;
$$\sum F_{\rm on2}=\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.02) }-\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.02) }$$
$$\sum F_{\rm on2}=\color{red}{\bf 0}\;{\rm N} $$
The force exerted on wire 3 by wire 1 is upward, while the force exerted on wire 3 by wire 2 is downward.
So that,
$$\sum F_{\rm on3}=F_{1on3}-F_{\rm 2on3}$$
$$\sum F_{\rm on3}=\dfrac{\mu_0 L_3I_1I_3}{2\pi d_{1\rightarrow 3}}-\dfrac{\mu_0 L_3I_3I_2}{2\pi d_{2\rightarrow 3}} $$
Plug the known;
$$\sum F_{\rm on3}=\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.04) }-\dfrac{(4\pi\times 10^{-7}) (0.5)(10)(10)}{2\pi (0.02) }$$
$$\sum F_{\rm on3}=(\color{red}{\bf -2.5\times 10^{-4}}\;{\rm N})\;\hat j$$
