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

Let the total displacement current be $I_d$.

For a point inside the region between the capacitor plates ($r < R$), the magnetic field is given by Ampere-Maxwell's law:
$B_{\text{in}} = \dfrac{\mu_0 I_d r}{2\pi R^2}$
At $r = R/2$:
$B(R/2) = \dfrac{\mu_0 I_d (R/2)}{2\pi R^2} = \dfrac{\mu_0 I_d}{4\pi R}$

For a point outside the region between the plates ($r > R$), the magnetic field is:
$B_{\text{out}} = \dfrac{\mu_0 I_d}{2\pi r}$
At $r = 2R$:
$B(2R) = \dfrac{\mu_0 I_d}{2\pi (2R)} = \dfrac{\mu_0 I_d}{4\pi R}$

The ratio of the magnetic fields is:
$\dfrac{B(R/2)}{B(2R)} = \dfrac{\dfrac{\mu_0 I_d}{4\pi R}}{\dfrac{\mu_0 I_d}{4\pi R}} = 1$

Thus, the required ratio is $1:1$.

Answer: $1:1$