Answer
$i=30$ $A$
Work Step by Step
Since the point is halfway between both wires, the magnetic field from one wire must equal the magnetic field from the other wire. Therefore, the magnetic field of one wire must be half of $300 \mu T$, or $150 \mu T$. Use the formula $$B=\frac{\mu_o i}{2\pi R}$$ Solving for $i$ yields $$i=\frac{2\pi B R}{\mu_o}$$ Substituting known values of $B=150 \times 10^{-6} T$, $R=4.0cm=0.040m$ (halfway between 0.0 cm and 8.0 cm), and $\mu_o=4\pi \times 10^{-7} H/m$ yields a current of $$i=\frac{2\pi (150\times 10^{-6}T)(0.040m)}{4\pi \times 10^{-7} H/m}=30 A$$
