A

To find the number of degrees of freedom of a gas, we can use the relation between the speed of sound ($c$), the density of the gas ($\rho$), and the number of degrees of freedom ($f$). The speed of sound in a gas is given by the formula:
$c = \sqrt{\frac{fRT}{M}}$
where:
- $f$ is the number of degrees of freedom,
- $R$ is the ideal gas constant (approximately $8.31 \, J \, mol^{-1} \, K^{-1}$),
- $T$ is the temperature in Kelvin (standard condition: $T = 273.15 \, K$),
- $M$ is the molar mass of the gas.

We first need to convert the density given in mg cm^{-3} to kg m^{-3}:
$\rho = 1.3 \, \text{mg cm}^{-3} = 1.3 \times 10^{-3} \, \text{g cm}^{-3} = 1.3 \times 10^{-3} \times 1000 \, kg m^{-3} = 1.3 \, kg m^{-3}$.

Now, we will find the molar mass of the gas ($M$). Using the ideal gas law:
$PV = nRT\Rightarrow n = \frac{PV}{RT}$
and substituting $n = \frac{\rho V}{M}$ gives $\rho = \frac{PM}{RT}$. Thus, we can rearrange for $M$:
$M = \frac{\rho RT}{P}$.
Assuming standard atmospheric pressure $P = 101325 \, Pa$:
$M = \frac{1.3 \times 10^{3} \cdot 8.31 \cdot 273.15}{101325} \Rightarrow M \approx 0.03 \text{ kg/mol}$ (approximately, this calculation yields a value around 30 g/mol).

Now substituting values into the speed of sound formula:
$330 = \sqrt{\frac{f \cdot 8.31 \cdot 273.15}{0.03}}$.
Squaring both sides gives:
$(330)^{2} = \frac{f \cdot 8.31 \cdot 273.15}{0.03}$.
Solving for $f$:
$f = \frac{(330)^{2} \cdot 0.03}{8.31 \cdot 273.15} \approx 5$.
The number of degrees of freedom for a monoatomic gas is 3, and for a diatomic gas, it is 5 (considering translational and rotational degrees of freedom).
Therefore, the gas likely has 5 degrees of freedom, which corresponds to diatomic gases.
Thus, the final answer is: 5.