A toroid has 1500 turns of wire and carries 0.60 A…

Step 1. The constant. $\dfrac{\mu_0 N I}{2\pi} = 2 \times 10^{-7}\ \text{T m/A} \times 1500 \times 0.60\ \text{A} = 1.8 \times 10^{-4}\ \text{T m}$.

Step 2. The inner edge. $B = \dfrac{1.8 \times 10^{-4}\ \text{T m}}{0.070\ \text{m}} = 2.57 \times 10^{-3}\ \text{T}$, that is $2.57$ mT.

Step 3. The outer edge. $B = \dfrac{1.8 \times 10^{-4}\ \text{T m}}{0.11\ \text{m}} = 1.64$ mT.

Step 4. The mean radius. $B = \dfrac{1.8 \times 10^{-4}\ \text{T m}}{0.090\ \text{m}} = 2.00$ mT. This equals $\mu_0 n I$ with $n = \dfrac{N}{2\pi r}$.

Step 5. At $14$ cm. This is outside the ring, so $B = 0$.