Two capacitors of capacitance 2 and 4 are connected in parallel....

The equivalent capacitance of the two capacitors in parallel is:
$C_p = 2 \mu\text{F} + 4 \mu\text{F} = 6 \mu\text{F}$

This combination $C_p$ is in series with $C_3$. The total equivalent capacitance $C_{eq}$ is given as $3 \mu\text{F}$.
For capacitors in series, the formula is:
$\dfrac{1}{C_{eq}} = \dfrac{1}{C_p} + \dfrac{1}{C_3}$

Substituting the known values:
$\dfrac{1}{3} = \dfrac{1}{6} + \dfrac{1}{C_3}$

$\dfrac{1}{C_3} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{2 - 1}{6} = \dfrac{1}{6}$

Therefore, $C_3 = 6 \mu\text{F}$.

Using the formula $C_{eq} = \dfrac{C_p \times C_3}{C_p + C_3}$ incorrectly to solve for $C_3$ as if it were the equivalent of $3$ and $6$ in series leads to the wrong answer of $2 \mu\text{F}$.

Answer: $6 \mu\text{F}$