Step 1: Use the series capacitance given as x:
The series net capacitance is $x = C/3$, so $C = 3x$. The parallel capacitance is $3C = 9x$.
Step 2: Energy at the same source:
$U\propto C_{eq}$ when $V$ is fixed. So $\dfrac{U_s}{U_p} = \dfrac{x}{9x} = \dfrac19$.
Step 3: Interpretation:
For the same voltage, parallel stores more energy because it has more capacitance. The factor 9 comes from $3\times3$, one 3 from the series reduction and one from the parallel increase.
Final Answer:
1:9, option (C).
\[ \boxed{1:9 \text{ (C)}} \]