Answer
$2.4\times10^{13}\,V/m\cdot s$
Work Step by Step
Given/Known: Radial distance from the central axis to the induced magnetic field $R=6.0\,mm=6.0\times10^{-3}\,m,$ Magnetic field $B=2.0\times10^{-7}\,T,$ Radius of the plates $r=3.0\,mm=3.0\times10^{-3}\,m,$ $\mu_{0}=4\pi\times10^{-7}\,Tm/A$ and $\varepsilon_{0}=8.85\times10^{-12}\,F/m$
According to Maxwell's law of induction,
$\oint \vec{B}\cdot d\vec{s}=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}=\mu_{0}\varepsilon_{0}\frac{d}{dt}(EA\cos0^{\circ})=\mu_{0}\varepsilon_{0}A\frac{dE}{dt}$
$\implies B\times2\pi R=\mu_{0}\varepsilon_{0}(\pi r^{2})\frac{dE}{dt}$
Or $\frac{dE}{dt}=\frac{B\times2\pi R}{\mu_{0}\varepsilon_{0}\times\pi r^{2}}=\frac{2BR}{\mu_{0}\varepsilon_{0}r^{2}}$
Result: $\frac{dE}{dt}=\frac{2\times2.0\times10^{-7}\,T\times6.0\times10^{-3}\,m}{4\pi\times10^{-7}\,Tm/A\times8.85\times10^{-12}\,F/m\times(3.0\times10^{-3}\,m)^{2}}=2.4\times10^{13}\,A/Fm$
$=2.4\times10^{13}\,V/m\cdot s$
