A long round bus bar, of radius 4.5 mm, carries 72 A spread evenly over its section…

Step 1. At $3.0$ mm, inside. $B = \dfrac{\mu_0 I r}{2\pi a^2}$. The top is $2 \times 10^{-7}\ \text{T m/A} \times 72\ \text{A} \times 3.0 \times 10^{-3}\ \text{m} = 4.32 \times 10^{-8}\ \text{T m}^2$. The bottom is $(4.5 \times 10^{-3}\ \text{m})^2 = 2.03 \times 10^{-5}\ \text{m}^2$. So $B = 2.13 \times 10^{-3}\ \text{T}$, that is $2.13$ mT.

Step 2. At the surface. $B = \dfrac{\mu_0 I}{2\pi a} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 72\ \text{A}}{4.5 \times 10^{-3}\ \text{m}} = 3.20 \times 10^{-3}\ \text{T}$, that is $3.20$ mT.

Step 3. At $13.5$ mm, outside. This is $3a$, so $B = \dfrac{3.20\ \text{mT}}{3} = 1.07$ mT.