c
Number of capacitors required in series $=\frac{3000}{500}=6.$
The capacitance of series combination $=(1 / 6)\, \mu \mathrm{F}.$
To obtain a capacitor of $2\, \mu \mathrm{F},$ we should use $12$ such combinations.
$\therefore $Total number of capacitors required $=12 \times 6=72$