Answer
$C = 5.0~\mu F$
Work Step by Step
In series, the charge $Q$ on each capacitor is equal.
Since the potential difference across the unknown capacitor is $6.0~V$, the potential difference across the $10~\mu F$ capacitor is $3.0~V$
We can find the capacitance $C$ of the unknown capacitor:
$Q = (10~\mu F)(3.0~V) = (C)(6.0~V)$
$C = \frac{(10~\mu F)(3.0~V)}{6.0~V}$
$C = 5.0~\mu F$
