A piece of wire 1.2 m long carries 3.0 A…

Step 1. One circle. $2\pi R = 1.2$ m, so $R = 0.191$ m. Then $B = \dfrac{\mu_0 I}{2R} = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 3.0\ \text{A}}{2 \times 0.191\ \text{m}} = 9.87 \times 10^{-6}\ \text{T}$, that is $9.87\ \mu\text{T}$.

Step 2. Three circles. Now $3 \times 2\pi R = 1.2$ m, so $R = 0.0637$ m. Then $B = \dfrac{\mu_0 N I}{2R} = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 3 \times 3.0\ \text{A}}{2 \times 0.0637\ \text{m}} = 8.88 \times 10^{-5}\ \text{T}$, that is $88.8\ \mu\text{T}$.

Step 3. The ratio. $\dfrac{88.8}{9.87} = 9$.