Solution:
Solution:
When there are two turns in the coil, then
$$l=2\times2\pi r_{1}$$ or $$r_{1}=\frac{l}{4\pi}$$
then $$B_{1}=\frac{\mu_{0} N_{1}I}{2r_{1}}=\frac{\mu_{0}\times2\times I}{2\times\left(l / 4\pi\right)}=\frac{\mu_{0}\,4\pi I}{l}$$
When there are four turns in the coil, then
$$l=4 \times2\pi r_{2}$$ or $$r_{2}=\frac{l}{8\pi}$$
Then $$B_{2}=\frac{\mu_{0}\, N_{2}I}{2r_{2}} = \frac{\mu_{0}\times4\times I}{2\times\left(l 8\pi\right)}=\frac{\mu_{0}\,16\pi I}{l}$$
$$\frac{B_{1}}{B_{2}}=\frac{4}{16}=\frac{1}{4}$$ or $$B_{2}=4B_{1}=4\times0.2\,T$$
$$=0.8\, T $$
