a b c d

d

Force due to first wire on middle wire is $F_1 = \frac{\mu_0 II_1}{6\pi}\times 25 = \frac{2500\mu_0}{2\pi}$   Force due to first wire on middle wire is $F_2 = \frac{\mu_0 II_2}{10\pi}\times 25 = \frac{1000\mu_0}{2\pi}$   Net Force F=F1​−F2​=​​$\frac{1500\mu_0}{2\pi}$=3×10−4N towards right