The equivalent capacitance between $C \& D$ capacitors of $2\, \mu \mathrm{F}, 5\, \mu \mathrm{F}$ and $5\, \mu \mathrm{F}$ are in parallel.

$\therefore \mathrm{C}_{\mathrm{CD}}=2+5+5=12 \,\mu \mathrm{F}$

($\because$ In parallel grouping $C_{e q}=C_{1}+C_{2}+\ldots .+C_{n})$

Similarly equivalent capacitance between $\mathrm{E}$ $\& \mathrm{BC}_{\mathrm{EB}}$

$=4+2=6\, \mu \mathrm{F}$

Now equivalent capacitance between $A \& B$

$\frac{1}{C_{e q}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$

$\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{12}{5}=24\, \mu \mathrm{F}$

($\because $ In series group-ing, $\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} +  \ldots  \ldots  + \frac{1}{{{C_n}}}$