Answer
See the detailed answer below.
Work Step by Step
As we see in the figures below, this is the cross-sectional area of the wire where the 3 dashed circles are the Ampere's paths.
$\bullet$ Applying Ampere’s law for the closed path; at $r_1$ which is when $r\lt R_1$
$$\oint \vec B\cdot d\vec s=\mu_0I_{\rm through}$$
where $I_{\rm through}=0$ here.
So that
$$\boxed{B_1=\color{red}{\bf 0}\;\rm T}$$
$\bullet\bullet$ Applying Ampere’s law for the closed path; at $r_2$ which is when $R_1\lt r\lt R_2$
$$\oint \vec B\cdot d\vec s=\mu_0I_{\rm through}$$
Recalling that the author told us that the current is uniformly distributed across the area
of the wire.
So, $I_{\rm through}=\dfrac{I}{A_{total}}A_{2}$ where $A_{total}$ is the cross-sectional area of the whole wire area while $A_1$ is the cross-section area of the part of the wire inside $r_2$.
$$B\oint \vec d s=\mu_0\dfrac{I}{A_{total}}A_{2}=\mu_0\dfrac{I}{\pi (R_2^2-R_1^2)} \pi(r ^2-R_1^2)$$
$$B (2\pi r) =\dfrac{\mu_0I(r ^2-R_1^2)}{ (R_2^2-R_1^2)} $$
$$\boxed{B_2 =\dfrac{\mu_0I(r ^2-R_1^2)}{ (2\pi r) (R_2^2-R_1^2)} }$$
$\bullet\bullet\bullet$ Applying Ampere’s law for the closed path; at $r_3$ which is when $ r_3\gt R_2$
$$\oint \vec B\cdot d\vec s=\mu_0I_{\rm through}$$
where $I_{\rm through}=I$ since it contains the whole area of the wire.
$$B(2\pi r)=\mu_0I $$
Hence,
$$\boxed{B_3=\dfrac{\mu_0I}{2\pi r}}$$
