Attempt:

$V_1 + V_2 = 440 $

$Q_1= Q_2 \implies k_1V_1C_1 = k_2V_2C_2$

$\implies V_2 = \dfrac{k_1C_1V_1}{k_2C_2}= \dfrac{9V_1}{35}$

$\implies V_1 = 350 V$

$E_1 = \dfrac{V_1}{d_1 k_1} = \dfrac{350}{0.07 \times 3}= \dfrac{5000}{3}$

But answers are: $E_1 = 5\times 10^4 V/m, E_2 = 3\times 10^4 V/m$

Note: I am using $E_1= \dfrac{V_1}{d_1k_1}$ because electric field in the presence of a dielectric is given by $\dfrac{E_o}{k}$ right? If I am misusing it please let me know. Also, what's the correct way to use this formula in this problem?

reshown Jul 18 by Reststack