Solution:

To solve this, we use the relationship from Charles's Law, which states that for a gas at constant pressure:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]

where:
- \( V_1 \) is the initial volume at temperature \( T_1 \),
- \( V_2 \) is the final volume at temperature \( T_2 \).

Given:
- Initial temperature \( T_1 = 0^\circ \text{C} = 273 \, \text{K} \),
- \( V_2 = 2V_1 \) (double the initial volume).

Now, substituting into Charles's Law:

\[
\frac{V_1}{273} = \frac{2V_1}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 273 = 546 \, \text{K}
\]

Converting \( T_2 \) back to Celsius:

\[
T_2 = 546 - 273 = 273^\circ \text{C}
\]

So, the required temperature is 273°C.