Final answer:
The total capacitance of the circuit containing three capacitors in parallel with capacitances of 0.02 µF, 0.05 µF, and 0.10 µF is 0.17 µF.
Step-by-step explanation:
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, the total capacitance would be 0.02 µF + 0.05 µF + 0.1 µF = 0.17 µF. Therefore, the correct answer is A. 170 μF.