Interpret the circuit

The usual arrangement for this standard problem is that $C_2=6\,\mu F$ and $C_3=12\,\mu F$ are in parallel, and that parallel combination is in series with $C_1=2\,\mu F$.

So first combine $C_2$ and $C_3$:

$C_{23}=C_2+C_3=6+12=18\,\mu F$

Series combination with $C_1$

Now $C_1=2\,\mu F$ is in series with $C_{23}=18\,\mu F$.

In a series combination, the charge is same on each series element.

Let the series charge be $Q$.

Then:

• charge on $C_1$ is $Q_1=Q$
• total charge on the parallel block is also $Q_{23}=Q$

Voltage across the parallel combination

The voltage across the parallel block is

$V_{23}=\frac{Q}{C_{23}}=\frac{Q}{18}$

Since $C_2$ and $C_3$ are in parallel, both have the same voltage $V_{23}$.

Charges on $C_2$ and $C_3$

Using $Q=CV$:

$Q_2=C_2V_{23}=6\cdot \frac{Q}{18}=\frac{Q}{3}$

$Q_3=C_3V_{23}=12\cdot \frac{Q}{18}=\frac{2Q}{3}$

Required ratio

Therefore,

$Q_1:Q_2:Q_3 = Q : \frac{Q}{3} : \frac{2Q}{3}$

Multiply by $3$:

$Q_1:Q_2:Q_3 = 3:1:2$

Compare with given options

The ratio $3:1:2$ is not present in the listed options.

Hence, based on the standard capacitor rules, the stored answer $\boxed{\text{C}}$ does not match the derived result.

Why stored answer may be incorrect

Option C is $1:2:2$, which would imply equal charges on $C_2$ and $C_3$, but parallel capacitors have equal voltage, not equal charge. Since $C_3$ is twice $C_2$, its charge should be twice that on $C_2$, not equal to it.

So the physically correct ratio for the described standard figure is:

$\boxed{3:1:2}$