Answer
$1.18\times10^{-19}\,T$
Work Step by Step
Given/Known: $\mu_{0}=4\pi\times10^{-7}\,T\cdot m/A,$ $\varepsilon_{0}=8.85\times10^{-12}\,F/m,$ $R=3.00\,cm=3.00\times10^{-2}\,m,$ $r=2.00\,cm=2.0\times10^{-2}\,m,$ $\Phi_{E}=(3.00\,mV\cdot m/s)t=(3.00\times10^{-3}\,V\cdot m/s)t$
Maxwell's law of induction states
$\oint \vec{B}\cdot d\vec{s}=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$
In this case,
$ B\times2\pi r=\mu_{0}\varepsilon_{0}\frac{d}{dt}(\Phi_{E}\times\frac{\pi r^{2}}{\pi R^{2}})$
Or $B=\frac{\mu_{0}\varepsilon_{0}}{2\pi }\times\frac{r}{R^{2}}\times\frac{d\Phi_{E}}{dt}$
Result:
$B=\frac{4\pi\times10^{-7}\,T\cdot m/A\times8.85\times10^{-12}\,F/m}{2\pi}\times\frac{2.0\times10^{-2}\,m}{(3.00\times10^{-2}\,m)^{2}}\times\frac{d}{dt}(3.00\times10^{-3}\,V\cdot m/s)=1.18\times10^{-19}\,T$
