Try doing the integration for one wire at distance $d$ from the x axis. When you have that answer you might be able to see the answer to the problem with many wires. It might be that the answer does not depend on the separation between the wires!
There is probably a simpler solution using Ampere's Law, which is the magnetic equivalent of Gauss' Law.
Consider a long thin rectangular circuit ABCDA around any one of the wires which carries current $I$. Side AB corresponds to the x axis in the diagram. By Ampere's Law the line integral of $B.dl$ around the circuit ABCDA is $\mu_0 I$, where $I$ is the enclosed current.
Now imagine that the rectangle is stretched further so that AB and CD become infinitely long. Although the magnetic field is parallel to BC and DA at these sides, it is very weak because these sides are so very far away from the wire, and they are very much shorter than the sides AB and CD. Almost all of the contribution to the line integral comes from AB and CD, and almost none from DA and BC. So in the limit as AB and CD become infinitely long, the line integral around circuit ABCDA is equal to the sum of the line integrals along AB and CD.
The wire is equidistant from AB and CD. By symmetry the line integrals along AB and CD are equal, so the infinite line integral along AB is $\frac12 \mu_0 I$. Note that this does not depend on the distance of the wire from the x axis AB.
The same reasoning can be applied to all of the currents in the diagram. The currents below the x axis point into the page, so these line integrals are also +ve in the direction AB. Using the superposition theorem, the line integral due to the magnetic fields of all the wires is $\frac12 \mu_0 $ times the sum of the currents.
All that remains for you to do is to sum an infinite geometric series of decreasing currents. Since this is a multiple choice question, you don't even have to do that. The line integral will be +ve and greater than $\frac12 \mu_0$ because the total current $I$ is greater than $1A$. There is only one option which meets these criteria.