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The formula to convert Celsius to Kelvin is $T_K = T_C + 273.15$.

$$T_1 = 20^\circ C + 273.15 = 293.15 \, \text{K}$$ $$T_2 = 30^\circ C + 273.15 = 303.15 \, \text{K}$$

Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature. The formula is:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

We are given $V_1 = 250 \, \text{cm}^3$, $T_1 = 293.15 \, \text{K}$, and $T_2 = 303.15 \, \text{K}$. We need to find $V_2$.

$$\frac{250 \, \text{cm}^3}{293.15 \, \text{K}} = \frac{V_2}{303.15 \, \text{K}}$$ $$V_2 = \frac{250 \, \text{cm}^3 \times 303.15 \, \text{K}}{293.15 \, \text{K}}$$ $$V_2 = \frac{75787.5}{293.15} \, \text{cm}^3$$ $$V_2 \approx 258.596 \, \text{cm}^3$$

The given values have 2 or 3 significant figures. Let's round to 3 significant figures.

$$V_2 \approx 259 \, \text{cm}^3$$

The volume which the gas sample would occupy at $30^\circ C$ is $\boxed{\text{259 cm}^3}$.