Answer: B
Explanation:
1) Apply Ampere's Law to the loop
[tex]\int{B*} \, ds = u_0I[/tex]
[tex]B(2\pi r)=u_0NI\\[/tex]
The loop has a circumference of 2pir.
The current enclosed is multiplied by N because there is N distinct loops.
[tex]B_1= \frac{u_0NI}{2\pi r}[/tex]
[tex]B_2=\frac{u_0(3N)I}{2\pi(0.5r) }[/tex]
[tex]B_2=6 B_1[/tex]
Answer is B