Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
At $r$ distance from the center of the wire,
We need the magnetic field inside the wire, so we need to use Ampere's law.
$$\oint Bds=\mu_0 I_{\rm through}$$
$$B(2\pi r)=\mu_0 I_{\rm through}$$
$$ B =\dfrac{ \mu_0I_{\rm through} }{2\pi r} \tag 1$$
$\bullet$ At $r\lt R_1$, where the charge density of the inner part is given by
$$\sigma=\dfrac{I}{A_{\rm inner}}=\dfrac{I}{\pi R_1^2}=\dfrac{I_{\rm through}}{\pi r^2}$$
Thus,
$$I_{\rm through}=\dfrac{Ir^2}{R_1^2}$$
Plug into (1)
$$ B =\dfrac{ \mu_0 Ir^2 }{2\pi rR_1^2} $$
$$ \boxed{B =\dfrac{ \mu_0 Ir }{2\pi R_1^2} }$$
$\bullet\bullet$ At $R_1\lt r\lt R_2$, $I_{\rm through}=I$, so plug into (1)
$$ \boxed{B =\dfrac{ \mu_0 I }{2\pi r} }$$
$\bullet\bullet\bullet$ At $r\gt R_2$, $I_{\rm through}=I-I=0$, so
$$ \boxed{B =\bf0\;\rm T }$$
$$\color{blue}{\bf [b]}$$
We have here 3 stages too, for the three boxed formulas above.
The first one is linear, the second is inversely proportional, and the third is just zero. See the graph below.
