Explanation

(a) The 10$\mu F$ and 6$\mu F$ capacitors are connected in parallel, hence resultant capacitance is
$\ \ \ \ \ \ \ \ C^{'} = 10\ \mu F + 6\ \mu F = 16\ \mu F\ \ \ $
This is connected in series with 4$\ \mu F\ \ $capacitor, hence effective capacitance is
$\ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{16} + \frac{1}{4} = \frac{20}{16 \times 4}$
$\Rightarrow \ \ C^{''} = \frac{64}{20} = 3.20\mu F$