Interpret the circuit condition

The capacitor is connected in a DC circuit. In the steady state, the capacitor behaves like an open circuit, so no current flows through the branch containing the capacitor.

Find the potential difference across the capacitor

From the resistor network in the given circuit, the two terminals across which the capacitor is connected have a potential difference of $V = 2.67\text{ V}$ (equivalently, about $\frac{8}{3}\text{ V}$).

Use the capacitor charge relation

Charge stored on a capacitor is $Q = CV$

Given, $C = 150\,\mu\text{F}$ and $V = \frac{8}{3}\text{ V}$

Therefore, $Q = 150 \times \frac{8}{3}\,\mu\text{C}$

$Q = 50 \times 8\,\mu\text{C}$

$Q = 400\,\mu\text{C}$

Final answer

$\boxed{400\,\mu\text{C}}$