Phy que 17.

Solution,
According to Ampere's circuital law, “the line integral of the magnetic field surrounding closed-loop equals to the number of times the algebraic sum of currents passing through the loop.”
$\oint B.dr={\mu }_o{I}_{encl}$
So, let us consider an Amperian loop passing through all the three loops as shown in the figure:

Thus the net current enclosed by the Amperian loop is:
I_en = 7i_o - 3i_o - i_o = 3i_{o
Hence,
$$\begin{gathered} {\int }_{-\infty }^{+\infty }B·dr={\mu }_o\times 3i_o=3{\mu }_o{i}_o \\ Thus, \\ x=3 \end{gathered}$$}