Answer
$\displaystyle \color{blue}{F_X(x)= \sum_{k=0}^x {4\choose k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4-k}, \ x = 0,1,2,3,4}.$
Work Step by Step
$\underline{\text{pdf of}\ X}$
Since the r.v. $X$ denotes the number of 6's in four ($n=4$) rolls of a die, the only possible values of $X$ are $x = 0, 1,2,3,4$.
Thus,
i) if we think of the outcome 6 as a "success" and an outcome that is not a 6 as a "failure",
ii) so that $X$ is the number of successes in $n=4$ independent trials (the four rolls of a die), the probability of success is $p=\dfrac{1}{6}$ (since there are 6 equally likely possible outcomes for each roll), and the probability of a failure is $q=1-p=\dfrac{5}{6}$,
by Theorem 3.2.1 (p. 104), the probability distribution of $X$ must be the binomial distribution having the form:
$\displaystyle P(X=x) = {4\choose x}\left(\frac{1}{6}\right)^x\left(\frac{5}{6}\right)^{4-x}, \ x = 0,1,2,3,4.$
Since the pdf of $X$ is $p_X(x) = P(X=x)$, then the pdf of $X$ is
$\displaystyle p_X(x) = {4\choose x}\left(\frac{1}{6}\right)^x\left(\frac{5}{6}\right)^{4-x}, \ x = 0,1,2,3,4.$
$\underline{\text{cdf of}\ X}$
$\begin{align*}
F_X(x) &= P(X\le x) & [\text{by Definition of a cdf}]\\
&= \sum_{k=0}^x P(X=k), x=0,1,2,3,4 & [\text{since}\ P(X=x) \ne 0\ \text{only for}\ x=0,1,2,\ldots,6]\\
&= \sum_{k=0}^x p_X(k), x=0,1,2,3,4 & [\text{Definition of a pdf}]\\
\color{blue}{F_X(x)} &\color{blue}{= \sum_{k=0}^x {4\choose k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4-k}, \ x = 0,1,2,3,4}. & [\text{see pdf of}\ X\ \text{above}]
\end{align*}$
