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(D) For a single circular loop of length $L$,the radius $R$ is given by $L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center is $B_L

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$\frac{1}{N^2}$

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(D) For a single circular loop of length $L$,the radius $R$ is given by $L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center is $B_L = \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.For a coil with $N$ turns,the length $L = N(2\pi R')$,so the radius $R' = \frac{L}{2\pi N} = \frac{R}{N}$.The magnetic field at the center of the coil is $B_C = \frac{N \mu_0 i}{2R'} = \frac{N \mu_0 i}{2(R/N)} = \frac{N^2 \mu_0 i}{2R}$.Taking the ratio,$\frac{B_L}{B_C} = \frac{\mu_0 i / 2R}{N^2 \mu_0 i / 2R} = \frac{1}{N^2}$.

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