Answer
$C_b > C_a=C_d>C_c$
Work Step by Step
We can find the capacitance of parallel capacitors by using
$C_{eq} = \displaystyle \Sigma_i^n$ $C_i = C_1+C_2+...+C_n$
and the capacitance of series capacitors using
$C_{eq} = \Sigma_i^n$ $\displaystyle \frac{1}{C_{i}} = \frac{1}{C_1} +\frac{1}{C_2} +...+\frac{1}{C_n}$
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The capacitance of A is just $5\mu$F since there's only one capacitor
The capacitance of B is $6\mu$F since they're connected in parallel and
$C_b = 3 + 3 = 6\mu$F
The capacitance of C is $1.5\mu$F since they're connected in series and
$\displaystyle \frac{1}{C_c} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
$\displaystyle C_c = \frac{3}{2}\mu$F
The capacitance of D is $5\mu$F. We must first find the capacitance of the capacitors on the right side which are in series.
$\displaystyle \frac{1}{C_{right}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
$C_{right} = 2\mu$F
Now D is in parallel:
$C_d = C_{left} + C_{right} = 3+2 = 5\mu$F
From largest to smallest capacitance, we can rank these as
$C_b > C_a=C_d>C_c$
