From the figure, the potential difference between the points  ‘x’ and ‘y’ is 

From diagram, in the branch  AxB, the current will be 0 in the steady state. Hence the current will flow only in the branch AyB.
The effective resistance experienced by the 24 V battery can we calculated as shown below
Effective resistance 2+4+6=12
By using ohm’s law we can get the current in the circuit, which is equal to 2A .
By using KVL, we can find the potential difference across  BA, which is equal to  24-2(2)=20 V
By using the formula  $V_1=\left(\frac{R_1}{R_1+R_2}\right)V,$
We get  $V_B-V_y=20\times \frac{6}{10}=12\text{\hspace{0.17em}V---(1)}$
By using formula  $V_1=\left(\frac{C_2}{C_1+C_2}\right)V=\left(\frac{{\displaystyle \frac{4}{3}}}{\frac{4}{3}+4}\right)20=\left(\frac{{\displaystyle \frac{4}{3}}}{\frac{16}{3}}\right)20=5V$
We get  $V_B-V_x=5\text{\hspace{0.17em}V----(2)}$
From above equations (1) & (2)   $V_x-V_y=7\text{\hspace{0.17em}V}$
Therefore, the correct answer is (A).