Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's imagine a Gaussian cylinder inside the given wire of two segments. The center of the cylinder is at the boundary between the two segments.
The cross-sectional area of this Gaussian cylinder is $A_g =\pi r^2$ where $r\leq R$ where $R$ is the radius of the original two-segment wire.
We know that the current is conserved, so
$$I_1=I_2=I$$
and since both segments have the same radius,
$$J_1=J_2=J $$
where $J=\sigma E$, hence,
$$J=\sigma_1 E_1=\sigma_2 E_2=\dfrac{I}{A}$$
Thus,
$$E_1=\dfrac{I}{A\sigma_1}\;\;\;\;\;\;{\rm and} \;\;\;\;\;\; E_2=\dfrac{I}{A\sigma_2}\tag 1$$
Recalling that the flux is given by
$$\Phi=\oint EdA $$
And for enclosed surfaces $\Phi=Q_{\rm enclosed}/\epsilon_0$
$$\Phi=E_2A_g -E_1A_g =\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
$$A( E_2 -E_1) =\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
Plug from (1),
$$A_g\left( \dfrac{I}{A\sigma_2}-\dfrac{I}{A\sigma_1}\right) =\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
where $Q_{\rm enclosed}=\eta A_g$ on the boundary between two segments of the Gaussian cylinder.
$$ \color{red}{\bf\not} A_g\left( \dfrac{I}{A\sigma_2}-\dfrac{I}{A\sigma_1}\right) =\dfrac{\eta \color{red}{\bf\not} A_g}{\epsilon_0}$$
Hence,
$$\boxed{\eta=\dfrac{\epsilon_0I}{A}\left( \dfrac{1}{ \sigma_2}-\dfrac{1}{ \sigma_1}\right) }$$
$$\color{blue}{\bf [b]}$$
From the previous boxed formula,
$$\eta=\dfrac{Q}{A}=\dfrac{\epsilon_0I}{A}\left( \dfrac{1}{ \sigma_2}-\dfrac{1}{ \sigma_1}\right)$$
Hence,
$$Q= \epsilon_0I \left( \dfrac{1}{ \sigma_{\rm iron}}-\dfrac{1}{ \sigma_{\rm copper}}\right)$$
Plug the known;
$$Q= (8.85\times 10^{-12})(5)\left( \dfrac{1}{ 1\times 10^7 }-\dfrac{1}{ 6\times 10^7}\right)$$
$$Q=\color{red}{\bf 3.69\times 10^{-18}}\;\rm C$$
