Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:

$$F = \frac{\mu_0 \cdot I_1 \cdot I_2 \cdot l}{2\pi \cdot r}$$

Force due to the Left Wire ($$I_1 = 30\text{ A}$$): 

$$F_1 = \frac{2 \times 10^{-7} \cdot 30 \cdot 10 \cdot 0.25}{0.03} = 5 \times 10^{-4}\text{ N (Right)}$$

Force due to the Right Wire ($$I_2 = 20\text{ A}$$):

$$F_2 = \frac{2 \times 10^{-7} \cdot 20 \cdot 10 \cdot 0.25}{0.05} = 2 \times 10^{-4}\text{ N (Left)}$$

$$F_{net} = F_1 (\text{Right}) - F_2 (\text{Left})$$

$$F_{net} = 5 \times 10^{-4}\text{ N} - 2 \times 10^{-4}\text{ N} = 3 \times 10^{-4}\text{ N}$$