A capacitor of 10 is connected in parallel with a series...

Let the equivalent capacitance of the two identical capacitors in series be $C_s$.

Since they are in series, $C_s = \dfrac{C \times C}{C + C} = \dfrac{C}{2}$.

This series combination is connected in parallel with a $10 \mu\text{F}$ capacitor. The equivalent capacitance of the entire arrangement is given as $14 \mu\text{F}$.

For a parallel combination, the equivalent capacitance is the sum of the individual capacitances:
$C_{\text{eq}} = 10 \mu\text{F} + C_s$

Substituting the given values:
$14 \mu\text{F} = 10 \mu\text{F} + \dfrac{C}{2}$

$\dfrac{C}{2} = 14 \mu\text{F} - 10 \mu\text{F} = 4 \mu\text{F}$

$C = 8 \mu\text{F}$

Answer: $8 \mu\text{F}$