(A) Equivalent Capacitance of the Circuit:
The given circuit has capacitors arranged in a combination of series and parallel connections. Let us simplify step by step:
1. The \( 6 \, \mu\mathrm{F} \) and \( 12 \, \mu\mathrm{F} \) capacitors are in series:
\[
\frac{1}{C_1} = \frac{1}{6} + \frac{1}{12} \quad \Rightarrow \quad C_1 = \frac{12}{3} = 4 \, \mu\mathrm{F}.
\]
2. The \( 6 \, \mu\mathrm{F} \) and \( 3 \, \mu\mathrm{F} \) capacitors are in series:
\[
\frac{1}{C_2} = \frac{1}{6} + \frac{1}{3} \quad \Rightarrow \quad C_2 = 2 \, \mu\mathrm{F}.
\]
3. The \( C_1 = 4 \, \mu\mathrm{F} \), \( 2 \, \mu\mathrm{F} \), and \( C_2 = 2 \, \mu\mathrm{F} \) are in parallel:
\[
C_{\text{eq}} = C_1 + C_2 + 2 \quad \Rightarrow \quad C_{\text{eq}} = 4 + 2 + 2 = 8 \, \mu\mathrm{F}.
\]
Thus, the equivalent capacitance of the circuit is:
\[
\boxed{8 \, \mu\mathrm{F}}.
\]
(B) Charge on \( 3 \, \mu\mathrm{F} \) and \( 2 \, \mu\mathrm{F} \) capacitors:
The total charge in the circuit is given by:
\[
Q_{\text{total}} = C_{\text{eq}} \cdot V = 8 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 80 \, \mu\mathrm{C}.
\]
For the \( 3 \, \mu\mathrm{F} \) capacitor, the voltage across \( C_2 = 2 \, \mu\mathrm{F} \) is the same as the voltage across the series combination of \( 6 \, \mu\mathrm{F} \) and \( 3 \, \mu\mathrm{F} \). Using charge conservation:
\[
Q = C \cdot V = 3 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 30 \, \mu\mathrm{C}.
\]
Similarly, for the \( 2 \, \mu\mathrm{F} \) capacitor:
\[
Q = C \cdot V = 2 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 20 \, \mu\mathrm{C}.
\]
Thus, the charges are:
\[
\boxed{30 \, \mu\mathrm{C} \text{ on the } 3 \, \mu\mathrm{F} \text{ capacitor, and } 20 \, \mu\mathrm{C} \text{ on the } 2 \, \mu\mathrm{F} \text{ capacitor.}}
\]