After a long time capacitor is charged and it acts like an open circuit. So, equivalent circuit is

So, current in circuit is

$I=\frac{6}{(1+2) \times 10^3}=2 \times 10^{-3} \,A$

Potential drop across $2 \,ks \Omega$ resistor is

$V_{A B}=I \cdot R_{A B}$

$=2 \times 10^3 \times 2 \times 10^3=4 \,V$

Clearly, potential drop across capacitor is also $4 \,V$.