Find the Current using Force/Length Between Two Wires

[sozo]

Homework Statement

Two long parallel wires are separated by .006 m. The current in one of the wires is twice the other current. If the magnitude of the force on a 3m length of one of the wires is equal to 8μN, what is the greater of the two currents?

The answer: .40 A

Homework Equations

F/L = (μ0)(Ia)(Ib) / 2∏d
μ0 = 4∏x10^-7
d = distance between the wires
Ia/b = values for the two currents

The Attempt at a Solution

I plugged everything in like so: (8x10^-6)/3 = (4∏x10^-7)(I)(2I)/2∏(.06)
From that, I divided the left side and the right, leaving me with:
2.67x10^-6 = 3.33x10^-6(2I^2)
Then I just solved for I by getting I by itself (.801/2 = I^2) and and taking the square root of (.801/2) to find I. At this point, I've found that I = .63A , but I don't know where to go from here. The answer should come out to be .40A.

 

[rude man]

The problem says d = 0.006m but you used d = 0.06m