Answer:
Total capacitance of the three capacitors in series = 0.4615uF
Explanation:
Take the inverse of the three capacitors (in this case you are working with µF so 10^-6) and add their inverse. The result is ~2.167*10^6. The inverse of this number is 0.4615 so the total capacitance of this circuit is 0.4615µF.
