A capacitor C_1 of 4 is connected in series with an unknown...

In a series combination, the charge $Q$ on both capacitors is the same.

The energy stored in a capacitor is given by $U = \dfrac{Q^2}{2C}$. This shows that for a constant charge, the energy stored is inversely proportional to the capacitance ($U \propto \dfrac{1}{C}$).

Given that the energy stored in $C_1$ is $4$ times the energy stored in $C_2$ ($U_1 = 4 U_2$), we can write:
$\dfrac{Q^2}{2C_1} = 4 \left(\dfrac{Q^2}{2C_2}\right)$

Simplifying this, we get:
$\dfrac{1}{C_1} = \dfrac{4}{C_2}$

Rearranging for $C_2$ gives:
$C_2 = 4 C_1$

Substituting the given value $C_1 = 4 \mu\text{F}$:
$C_2 = 4 \times 4 \mu\text{F} = 16 \mu\text{F}$

Answer: $16 \mu\text{F}$