A circular parallel plate capacitor of radius R is being charged...

According to the Ampere-Maxwell law, the line integral of the magnetic field along a circular Amperian loop of radius $r$ is $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,\text{enclosed}}$.

For a point inside the capacitor ($r < R$), the enclosed displacement current is proportional to the area: $I_{d,\text{enclosed}} = I_d \left(\dfrac{\pi r^2}{\pi R^2}\right)$.
Therefore, $B(2\pi r) = \mu_0 I_d \dfrac{r^2}{R^2} \Rightarrow B = \left(\dfrac{\mu_0 I_d}{2\pi R^2}\right) r$.
This shows that $B$ increases linearly with $r$ for $r < R$.

For a point outside the capacitor ($r > R$), the enclosed displacement current is the total displacement current $I_d$.
Therefore, $B(2\pi r) = \mu_0 I_d \Rightarrow B = \dfrac{\mu_0 I_d}{2\pi r}$.
This shows that $B$ decreases as $1/r$ for $r > R$.

Hence, the magnetic field increases linearly with $r$ inside the plates and decreases as $1/r$ outside the plates.

Answer: $B$ increases linearly with $r$ for $r < R$, and decreases as $1/r$ for $r > R$.