Answer
See the detailed answer below.
Work Step by Step
We know that the electric field of an infinite plane of charge is given by
$$E=\dfrac{\eta}{2\epsilon_0}$$
Now we know that the lower infinite plane will polarize the conductor, as we see in the figure below,
So the conductor now has two charge densities on the upper and lower surfaces.
We know that in Region 2, inside the conductor, the electric field had to be zero.
Hence,
$$E_{net,2}=E_C-E_A-E_C=\dfrac{\eta}{2\epsilon_0}-\dfrac{\eta_1}{2\epsilon_0}-\dfrac{\eta_1}{2\epsilon_0}=0$$
So,
$$\eta-2\eta_1=0$$
hence,
$$\boxed{\eta_1=\frac{1}{2}\eta}$$
We are dealing with magnitudes, not signs.
From the figure below, we can see that,
$\bullet$ At region 1:
$$E_{net,1}=E_C+E_A-E_B=\dfrac{\eta}{2\epsilon_0}+\dfrac{\eta}{4\epsilon_0}-\dfrac{\eta}{4\epsilon_0}$$
$$\boxed{\vec E_{net,1}=\dfrac{\eta}{2\epsilon_0}\;\hat j}$$
$\bullet\bullet$ At region 2: inside the conductor,
$$E_{net,2}=E_C-E_A-E_B=\dfrac{\eta}{2\epsilon_0}-\dfrac{\eta}{4\epsilon_0}-\dfrac{\eta}{4\epsilon_0}$$
$$\boxed{\vec E_{net,2}=\bf 0 \;\rm N/C}$$
$\bullet\bullet\bullet$ At region 3:
$$E_{net,3}=E_C+E_B-E_A=\dfrac{\eta}{2\epsilon_0}+\dfrac{\eta}{4\epsilon_0}-\dfrac{\eta}{4\epsilon_0}$$
$$\boxed{\vec E_{net,3}= \dfrac{\eta}{2\epsilon_0}\hat j}$$
$\bullet\bullet\bullet\bullet$ At region 4:
$$E_{net,4}=E_B-E_A-E_C=\dfrac{\eta}{4\epsilon_0}-\dfrac{\eta}{4\epsilon_0}-\dfrac{\eta}{2\epsilon_0}$$
$$\boxed{\vec E_{net,4}= -\dfrac{\eta}{2\epsilon_0}\hat j}$$
