Answer
${\bf 8.66}\;\rm T/s$
Work Step by Step
We have here a loop coil and a changing flux, so we must have an induced current.
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|=IR$$
where $\phi =AB\cos\theta$ and here $A$ is constant and $\theta=0^\circ$, so that
$$ A \left|\dfrac{dB}{dt}\right|=IR$$
And for $N$ turns.
$$ \pi r^2N \left|\dfrac{dB}{dt}\right|=IR$$
Hence,
$$ \left|\dfrac{dB}{dt}\right|=\dfrac{IR}{ \pi r_{\rm coil}^2N}\tag 1$$
Now we need to find $R$ of the whole wire where $R$ is given by
$$R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi r^2}$$
where $L$ is the whole length of the wire that is given by $L=2\pi r_{\rm coil} N$
$$R =\dfrac{2\color{red}{\bf\not} \pi r_{\rm coil} N\rho_{\rm Cu} }{\color{red}{\bf\not} \pi r_{\rm wire}^{ 2}}$$
$$R =\dfrac{2 r_{\rm coil} N\rho_{\rm Cu} }{ r^2_{\rm wire} }$$
Plug into (1),
$$ \left|\dfrac{dB}{dt}\right|=\dfrac{2\color{red}{\bf\not} r_{\rm coil} \color{red}{\bf\not} N\rho_{\rm Cu} I }{ \pi r_{\rm coil}^{\color{red}{\bf\not} 2}r^2_{\rm wire} \color{red}{\bf\not} N} $$
$$ \left|\dfrac{dB}{dt}\right|=\dfrac{2 \rho_{\rm Cu} I }{ \pi r_{\rm coil} r^2_{\rm wire} } $$
Plug the known;
$$ \left|\dfrac{dB}{dt}\right|=\dfrac{2 (1.7\times 10^{-8})(2) }{ \pi (4\times 10^{-2}) (0.25\times 10^{-3})^2 } $$
$$ \left|\dfrac{dB}{dt}\right|=\color{red}{\bf 8.66}\;\rm T/s$$
