Answer
1.4 T/s
Work Step by Step
Faraday's law of induction states
$\varepsilon=-\frac{d\Phi_{B}}{dt}$
$\vec{B}$ is perpendicular to $\vec{A}$ and $\vec{B}$ is uniform. Therefore, $\Phi_{B}=BA$ which gives
$\varepsilon=-\frac{d(BA)}{dt}=-A\frac{dB}{dt}$
$\implies |\frac{dB}{dt}|=\frac{\varepsilon}{A}=\frac{iR}{A}$
Now, $i=10\,A$ (given), $A=\pi\times(\frac{d}{2})^{2}=\pi(5\times10^{-2}\,m)^{2}=7.85\times10^{-3}\,m^{2}$ and
$R=\frac{\rho L}{Area}=\frac{\rho\times\pi d}{\pi (D/2)^{2}}=\frac{\rho\times d}{(D/2)^{2}}=\frac{1.69\times10^{-8}\Omega\cdot m\times10\times10^{-2}\,m}{(1.25\times10^{-3}\,m)^{2}}$
$=1.08\times10^{-3}\,\Omega$
Substituting the values of i, R and A, we get
$|\frac{dB}{dt}|=\frac{10\,A\times1.08\times10^{-3}\,\Omega}{7.85\times10^{-3}\,m^{2}}=1.4\,T/s$
