Assume the electric field inside the capacitor is uniform and use the Ampere-Maxwell law.
    
b) The induced magnetic field inside a charging capacitor is 
 $B=\frac{{\mu }_0}{2\pi }\frac{r}{R^2}\frac{dQ}{dt}=\frac{{\mu }_0}{2\pi }\frac{r}{R^2}{I}_{wire}$
Where we used $I_{wire}=dQ/dt$  as the actual current in  the wire leading to the capacitor. At  $t=2.0mm$ ,
 $B=\frac{(4\pi \times {10}^{-7}Tm/A)(2.0\times {10}^{-3}m)}{2\pi {(5.0\times {10}^{-3}m)}^2}\left(10A\right)$
 $=1.6\times {10}^{-4}T$