The induced emf in the ring is given by Faraday's law: $\mathcal{E} = \left| -\frac{d\Phi}{dt} \right| = \pi a^2 \frac{dB}{dt} = \pi a^2 b$.
The induced non-conservative electric field $E_{in}$ is uniform along the ring. By symmetry, $\oint \vec{E}_{in} \cdot d\vec{l} = \mathcal{E} \implies E_{in}(2\pi a) = \pi a^2 b \implies E_{in} = \frac{ab}{2}$. Thus, C is correct.
Let the resistance of the lower-resistance half be $R_0$; then the higher-resistance half is $\eta R_0$. Total resistance $R = (\eta + 1)R_0$.
The induced current in the ring is $I = \frac{\mathcal{E}}{R} = \frac{\pi a^2 b}{(\eta + 1)R_0}$.
By Ohm's law in local form, the net electric field inside a conductor is proportional to the current density. The net electric field $E$ in a segment of length $l$ and resistance $R_{seg}$ is $E = \frac{V_{drop}}{l} = \frac{I R_{seg}}{l}$.
For the higher resistance half (length $\pi a$): $E_1 = \frac{I (\eta R_0)}{\pi a} = \frac{\pi a^2 b \eta R_0}{(\eta + 1)R_0 \pi a} = \frac{\eta a b}{\eta + 1}$. Thus, A is correct.
For the lower resistance half: $E_2 = \frac{I (R_0)}{\pi a} = \frac{\pi a^2 b R_0}{(\eta + 1)R_0 \pi a} = \frac{a b}{\eta + 1}$. Thus, B is correct.
The net electric field is the vector sum of the non-conservative induced field $\vec{E}_{in}$ and the conservative electrostatic field $\vec{E}_c$ created by surface charges accumulated at the junctions: $\vec{E}_{net} = \vec{E}_{in} + \vec{E}_c$.
Taking the direction of current as positive, for the first half: $E_1 = E_{in} + E_{c1} \implies E_{c1} = E_1 - E_{in} = \frac{\eta ab}{\eta+1} - \frac{ab}{2} = \frac{ab(2\eta - \eta - 1)}{2(\eta+1)} = \frac{ab(\eta-1)}{2(\eta+1)}$.
For the second half: $E_2 = E_{in} + E_{c2} \implies E_{c2} = E_2 - E_{in} = \frac{ab}{\eta+1} - \frac{ab}{2} = \frac{ab(2 - \eta - 1)}{2(\eta+1)} = -\frac{ab(\eta-1)}{2(\eta+1)}$.
The magnitudes of the conservative field in both halves are equal to $\frac{ab(\eta-1)}{2(\eta+1)}$. Thus, D is correct.

Induced Electric Field