Determine the charges on the four capacitors having capacitances 1...

In steady state, capacitors act as open circuits, so no direct current flows through the branches containing only capacitors.

Let the potential of the negative terminal of the $6\text{ V}$ battery be $0\text{ V}$ and the positive terminal be $6\text{ V}$. The left vertical wire is at $6\text{ V}$ and the right vertical wire is at $0\text{ V}$.

For the branch with $1\text{ }\Omega$ and $2\text{ }\Omega$ resistors, they are in series. The current through this branch is:
$I_1 = \dfrac{6}{1 + 2} = 2\text{ A}$

The potential at the node between the $1\text{ }\Omega$ and $2\text{ }\Omega$ resistors is:
$V_1 = 6 - (2 \times 1) = 4\text{ V}$

This node is connected to the node between the $1\text{ }\mu\text{F}$ and $2\text{ }\mu\text{F}$ capacitors, so the potential between these capacitors is also $4\text{ V}$.

The potential difference across the $1\text{ }\mu\text{F}$ capacitor is $6 - 4 = 2\text{ V}$. Its charge is:
$q_1 = 1\text{ }\mu\text{F} \times 2\text{ V} = 2\text{ }\mu\text{C}$

The potential difference across the $2\text{ }\mu\text{F}$ capacitor is $4 - 0 = 4\text{ V}$. Its charge is:
$q_2 = 2\text{ }\mu\text{F} \times 4\text{ V} = 8\text{ }\mu\text{C}$

For the branch with two $3\text{ }\Omega$ resistors, they are in series. The current through this branch is:
$I_2 = \dfrac{6}{3 + 3} = 1\text{ A}$

The potential at the node between the two $3\text{ }\Omega$ resistors is:
$V_2 = 6 - (1 \times 3) = 3\text{ V}$

This node is connected to the node between the $3\text{ }\mu\text{F}$ and $4\text{ }\mu\text{F}$ capacitors, so the potential between these capacitors is also $3\text{ V}$.

The potential difference across the $3\text{ }\mu\text{F}$ capacitor is $6 - 3 = 3\text{ V}$. Its charge is:
$q_3 = 3\text{ }\mu\text{F} \times 3\text{ V} = 9\text{ }\mu\text{C}$

The potential difference across the $4\text{ }\mu\text{F}$ capacitor is $3 - 0 = 3\text{ V}$. Its charge is:
$q_4 = 4\text{ }\mu\text{F} \times 3\text{ V} = 12\text{ }\mu\text{C}$

The charges on the four capacitors are $2\text{ }\mu\text{C}$, $8\text{ }\mu\text{C}$, $9\text{ }\mu\text{C}$, and $12\text{ }\mu\text{C}$ respectively.

Answer: $2 \text{ }\mu\text{C}$, $8 \text{ }\mu\text{C}$, $9 \text{ }\mu\text{C}$, and $12 \text{ }\mu\text{C}$