The answer to this question would be:C. 22mf
In this question, there are two capacitors that connected in parallel. The formula to count the capacitor is the opposite of the formula used for the resistor. If constructed in series, the capacitor will have reduced capacitance. For parallel condition the equation would be:
Ctotal= C1+C2
C total= 2 mf+ 20mf= 22mf
