A flat circular coil has 60 turns and a radius of 9.0 cm…

Step 1. The centre. $B_0 = \dfrac{\mu_0 N I}{2R} = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 60 \times 0.45\ \text{A}}{2 \times 0.090\ \text{m}}$. This gives $1.885 \times 10^{-4}\ \text{T}$, that is $188\ \mu\text{T}$.

Step 2. At $12$ cm. The distance from the coil is $\sqrt{9^2 + 12^2} = 15$ cm. $B = B_0\left(\dfrac{R}{15\ \text{cm}}\right)^3 = 188\ \mu\text{T} \times 0.216 = 40.7\ \mu\text{T}$.

Step 3. Half the field. We want $\left(\dfrac{R^2}{R^2 + x^2}\right)^{3/2} = \dfrac{1}{2}$. So $R^2 + x^2 = 2^{2/3}R^2 = 1.587R^2$, and $x = 0.766R = 0.766 \times 9.0\ \text{cm} = 6.90$ cm.