The stored charge in the capacitor in steady state of the following...

In steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the $100\ \mu\text{F}$ capacitor. Consequently, no current flows through the $10\ \Omega$ resistor in series with it on the top right.

The voltage across the capacitor will be equal to the voltage across the $4\ \Omega$ vertical resistor.

Let's simplify the circuit to find this voltage. The $12\ \Omega$ resistor on the left is in parallel with the $12\ \text{V}$ battery, so the voltage across the rest of the circuit remains $12\ \text{V}$.

For the rightmost part of the active circuit, the $4\ \Omega$ top resistor, the $4\ \Omega$ vertical resistor, and the $2\ \Omega$ bottom resistor are in series. Their equivalent resistance is:
$R_{\text{right}} = 4\ \Omega + 4\ \Omega + 2\ \Omega = 10\ \Omega$

This $10\ \Omega$ equivalent resistance is in parallel with the $10\ \Omega$ vertical resistor in the middle. The equivalent resistance of this parallel combination is:
$R_{\text{mid}} = \dfrac{10 \times 10}{10 + 10} = 5\ \Omega$

Now, the total resistance of the main circuit connected to the $12\ \text{V}$ source is the sum of the $5\ \Omega$ top-left resistor, the $5\ \Omega$ equivalent middle resistance, and the $2\ \Omega$ bottom-left resistor:
$R_{\text{total}} = 5\ \Omega + 5\ \Omega + 2\ \Omega = 12\ \Omega$

The total current flowing from the $12\ \text{V}$ source into this main branch is:
$I = \dfrac{12\ \text{V}}{12\ \Omega} = 1\ \text{A}$

The voltage across the parallel combination (the middle section) is:
$V_{\text{mid}} = I \times R_{\text{mid}} = 1\ \text{A} \times 5\ \Omega = 5\ \text{V}$

This $5\ \text{V}$ is applied across the right branch which has a total resistance of $10\ \Omega$. The current through this branch is:
$I_{\text{right}} = \dfrac{5\ \text{V}}{10\ \Omega} = 0.5\ \text{A}$

The voltage across the $4\ \Omega$ vertical resistor is:
$V_{4\Omega} = I_{\text{right}} \times 4\ \Omega = 0.5\ \text{A} \times 4\ \Omega = 2\ \text{V}$

Since the capacitor is in parallel with this $4\ \Omega$ resistor (with no voltage drop across the $10\ \Omega$ resistor in its branch), the voltage across the capacitor is also $2\ \text{V}$.

The stored charge in the capacitor is:
$Q = C \times V = 100\ \mu\text{F} \times 2\ \text{V} = 200\ \mu\text{C}$

Answer: $200$