Let $X$ be the number of 6's in 300 rolls of the die.
If the die is fair, in the sense that the probability of a 6
on any one roll is $1/6,$ then $X \sim Binom(300, 1/6$ and $E(X) = 300(1/6) = 50.$
You observed $X = 60,$ which is more than expected. The
question is whether it is remarkable to get $X \ge 60$ if $X \sim Binom(300,1/6).$
Under that assumption it is easy to find $P(X \ge 60) = 0.073 ,$ using R statistical software. Or you can use the normal approximation to get
a close enough approximation.
1 - pbinom(59, 300, 1/6)
This is a small probability, that may cast some doubt on the assumption
that $X \sim Binom(300, 1/6),$ but not so small that most statisticians
and researchers would be willing to declare the die as 'unfair.'
I will leave it to you to put this into the format of a test of hypothesis
(as specified in your text),
with the value of the test statistic, and the P-value. How big would $X$
have to be ('critical value') in order to reject the null hypothesis at the 5% level of significance? (You might get a different critical value using an
exact binomial computation than with a normal approximation.)
I should say that $n = 300$ rolls of the die (although perhaps tedious
to perform) do not give a lot of information about the die. Based on 60 sixes out of 300, an Agresti 95% CI for the true probability of getting a 6 is fairly wide--about $(0.16,0.35).$
be best to have the counts, out of 300, for each of the six faces of
the die, and to do a chi-squared goodness-of-fit test of the null hypothesis
that each face has probability $1/6$ of showing on any one roll.
Notes: (a) It is worth pointing out that a die can be seriously biased in many
ways and still show 6's with the frequency expected of a fair die.
(b) It also possible to bias a die so that it is very unfair with respect
to a few faces and still show an average of 3.5 over the long run--and
show 6's a sixth of the time over the long run. All of this points to
the fact that keeping totals of all six faces would be the optimal way to
collect the data, but that is not the problem you were given to solve.
(c) Under the assumption that $X \sim Binom(300, 1/6),$ the formula
you use gives $Var(X) = 300(1/6)(5/6) = 41.667,$ so $SD(X) = 6.455$ is
indeed correct. The figure below shows that PDF of $Binom(300, 1/6)$
together with the PDF (red curve) of the approximating normal distribution $Norm(\mu = 50, \sigma = 6.455).$