Solution:

Two capacitors of \(2\text{ }\mu\text{F}\) in series give \(1\text{ }\mu\text{F}\). When this is connected in parallel with the third \(2\text{ }\mu\text{F}\) capacitor, the equivalent capacitance is \(1\text{ }\mu\text{F} + 2\text{ }\mu\text{F} = 3\text{ }\mu\text{F}\).