A long hollow pipe has an inner radius of 20 mm and an outer radius of 30 mm…

Step 1. At $10$ mm. The circle is in the hollow part. No current goes through it, so $B = 0$.

Step 2. At $25$ mm, in the wall. The metal inside the circle has an area in proportion to $r^2 - a^2 = 25^2 - 20^2 = 225$ (in mm$^2$ divided by $\pi$). The whole wall has $b^2 - a^2 = 500$. So $I_{enc} = 85\ \text{A} \times \dfrac{225}{500} = 38.25$ A. Then $B = \dfrac{\mu_0 I_{enc}}{2\pi r} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 38.25\ \text{A}}{0.025\ \text{m}} = 3.06 \times 10^{-4}\ \text{T}$, that is $306\ \mu\text{T}$.

Step 3. At $60$ mm, outside. All the current goes through the circle. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 85\ \text{A}}{0.060\ \text{m}} = 2.83 \times 10^{-4}\ \text{T}$, that is $283\ \mu\text{T}$.