The system can be thought of as a combination of three wires, each having a circular cross section . One of these, of radius #a# carries the same current density as our conductor, but now the current is spread over the entire circular cross section (not just over the region shaded in the picture). The other two are wires where the cavities are in the problem - two wires of diameter #a# carrying the same current density but in the opposite direction (into the page).
The new current in each of the smaller wires must be #I/2# each (the area of each smaller wire is half that of the original conductor.), while that of the larger wire is #2I#.
Now, the magnetic field at a distance #r# from the axis of a infinite straight wire carrying a current #I# is directed tangentially and its magnitude is given by
#B= {mu_0 I}/{2pi r}#
This works as long as the current is distributed uniformly through the cross-section of a wire with a circular cross section.
Point #P_1#
At point #P_1# the magnetic field due to the wire of radius #a# is directed to the left, and it has a magnitude of
#{mu_0 2I}/{2 pi r} = mu_0/pi I/r#
The point #P_1# is at distances of #r-a/2# and #r+a/2# from the centers of the two smaller wires. The two wires will both produce magnetic fields directed to the right, with magnitudes of
#{mu_0 I/2}/{2 pi (r-a/2)}# and #{mu_0 I/2}/{2 pi (r+a/2)}#
Thus, the net magnetic field at #P_1# due to our conductor is
# mu_0/pi I/r -{mu_0 I/2}/{2 pi (r-a/2)}-{mu_0 I/2}/{2 pi (r-+/2)} #
#qquad = mu_0/pi I/r (1-1/4(1/{1-a/{2r}}+1/{1+a/{2r}}) )#
#qquad = mu_0/pi I/r(1-1/{2(1-a^2/{4r^2})})#
#qquad = {mu_0 I}/{2pi r} (1-a^2/(2r^2))/(1-a^2/(4r^2))#
Point #P_2#
The field at point #P_2# due to the bigger wire is, again, #mu_0/pi I/r##, but this times to the north (i.e, vertically up).
The distance of the centers of the smaller wires from #P_2# is given by #sqrt{r^2+a^2/4}#. The magnitudes of the magnetic fields they both produce at #P_2# are both equal to #{mu_0 I/2}/{2pi sqrt{r^2+a^2/4}}#. Note, though, that they are both directed tangentially with respect to the respective centers - so that they are not in the same direction as in the previous case. However, it is easy to see that when you add the magnetic fields due to both the smaller wires up, you get their resultant in the north-south direction. This sum is
#2times {mu_0 I/2}/{2pi sqrt{r^2+a^2/4}}times r/sqrt{r^2+a^2/4} = {mu_0I}/{2pi r} 1/{1+a^2/{4r^2}}#
to the south. (The factor #r/sqrt{r^2+a^2/4} # is the cosine of the angle between the two fields and the north-south direction).
Thus. the net magnetic field at the point #P_2# is
#mu_0/pi I/r- {mu_0I}/{2pi r} 1/{1+a^2/{4r^2}} = {mu_0I}/{2pi r} {1+a^2/{2r^2}} /{1+a^2/{4r^2}} #
to the north.
