Between points E and D :
12μF and 6μF are connected in series, So replace them with a single capacitor C′
∴ C′=12×612+6=4μF
Now this 4μF is connected in parallel with another 4μF capacitor.
Equivalent capacitance between E and D CED=4μF+4μF=8μF
Now this 8μF is connected in series with another 1μF capacitor.
Equivalent capacitance between R and D CRD=1×81+8=89μF
Between points P and Q :
Equivalent capacitance between points P and Q, CPQ=2μF+2μF=4μF
This 4μF is connected in series with another 8μF capacitor.
Equivalent capacitance between R and Q CRQ=4×84+8=83μF
Equivalent capacitance between R and B CRB=83μF+89μF=329μF
Equivalent capacitance between A and B CAB=C×329C+329
∴ 1μF=329CC+329 ⟹C=3223μF