Solution :
Capacitance of each of the three capacitors, $C = 9\; pF $
Equivalent capacitance $(C_{eq})$ of the combination of the capacitors is given by the relation,
$\large\frac{1}{C_{eq}} =\frac{1}{C}+\frac{1}{C} +\frac{1}{C} =\frac{3}{C} =\frac{3}{9}=\frac{1}{3}$
=> $\large\frac{1}{C_{eq}} =\frac{1}{3} => $$C_{eq}=3 \mu F$
Therefore, total capacitance of the combination is $3 \mu F$.