Why is ans d?? In amperes law I is enclosed current but here i1and i2 is not enclosed

Dear Student,
Consider the amperian loop ABCA, then from Ampere's circuital law:
$\oint B.dl = {\mu }_o(i_1+i_3)$
Since, i₁ and i₃ are both in the direction of the area vector for the surface ABCA.

Consider the Amperian loop ACDA, then from Ampere's circuital law:
$\oint B.dl = {\mu }_o(i_2-i_3)$
Since, i₂ is in the direction and i₃ is opposite to the direction of the area vector for the surface ACDA .
Thus,
For the entire loop ABCDA:
$\oint B.dl = {\mu }_o(i_1+i_3)+{\mu }_o(i_2-i_3)={\mu }_o(i_1+i_2)$.
Ans (D)

Regards