If $\mu$ and $\sigma^{2}$ are the mean and variance of a random variable $X$ whose probability mass function is given by $P(X=x) = \binom{6}{x} \left(\frac{1}{3}\right)^{x} \left(\frac{2}{3}\right)^{6-x}$ for $x = 0, 1, 2, \ldots, 6$,then the value of $2\mu + 12\sigma^{2}$ is:
- A$4$
- B$8$
- C$20$
- D$16$
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