In steady state (af ter the capacitors are fully charged), no current flows through the branches containing capacitors. So, in steady state, given circuit is equivalent to circuit shown below.

Now, current in circuit is

$i=\frac{E}{R_{\text {total }}}=\frac{6}{3 \times 10^3}=2 \times 10^{-3} \,A$

Now, potential drop across $1 \mu F$ and $2 \,\mu F$ capacitors = potential drop across $2 \,k \Omega$ resistance

$=i R=2 \times 10^{-3} \times 2 \times 10^3=4 \,V$

So, charges stored in capacitors are

$Q_1=C_1 V=1 \times 10^{-6} \times 4=4 \,\mu C$

$\text { and } \quad Q_2=8 \,\mu C$