Somebody please help again

[CartoonKid]

http://www.webassign.net/pse/p30-21alt.gif

Figure P30.23 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.20 A out of the monitor, and the current in the outer conductor is I2 = 2.88 A into the monitor. Determine the magnitude and direction of the magnetic field at point b.

My working is:
B_1 = \frac{\mu_{0}I_1}{2\Pi(3\times10^{-3})}
B_2 = \frac{\mu_{0}I_2}{2\Pi(1\times10^{-3})}
B_2 - B_1 = 496\mu T
However, my answer is wrong. Somebody please help me.
The hint given after my wrong submission is:
Apply Ampere's Law and consider the currents inside the Amperian loop only.
Can somebody explain briefly to me, what is the main idea of Ampere's Law and suggest the approach we can use everytime we deal with Ampere's Law question.
Thank you very much.

[Andrew Mason]

My working is:
B_1 = \frac{\mu_{0}I_1}{2\Pi(3\times10^{-3})}
B_2 = \frac{\mu_{0}I_2}{2\Pi(1\times10^{-3})}
B_2 - B_1 = 496\mu T
However, my answer is wrong. Somebody please help me.
The hint given after my wrong submission is:
Apply Ampere's Law and consider the currents inside the Amperian loop only.You are not interested in B at point a. The I in Ampere's law is "enclosed current". At b, the enclosed current is the vector sum of the enclosed currents, ie. I2 - I1. So:

B_b = \frac{\mu_{0}(I_2-I_1)}{2\pi r_b}

AM