Answer
$1.06\times10^{-19}\,T$
Work Step by Step
Given/Known: $\mu_{0}=4\pi\times10^{-7}\,T\,m/A,$ $\varepsilon_{0}=8.85\times10^{-12}\,F/m,$ $r=5.00\times10^{-2}\,m,$ $\Phi_{E}=(3.00\times10^{-3}\,V\cdot m/s)t$
According to Maxwell's law of induction,
$\oint\vec{B}\cdot d\vec{s}=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ which gives
$B=\frac{\mu_{0}\varepsilon_{0}}{2\pi r}\times\frac{d\Phi_{E}}{dt}$
Result: $B=\frac{4\pi\times10^{-7}\,T\,m/A\times8.85\times10^{-12}\,F/m}{2\pi \times5.00\times10^{-2}\,m}\times3.00\times10^{-3}\,V\cdot m/s$
$=1.06\times10^{-19}\,T$
