Statistics Hypothesis Testing finding out test stat and critical value 
For a) 
$$z =\frac{ \bar{x} - \mu }{ \frac{\sigma }{ \sqrt n}}$$
I have deciphered that sample mean is $$\frac{20 + 23 + 21 + 22}{ 4} = 21.5$$
I came up with $1.29099..$ for Standard Deviation.
Sample size is $4$ since $4$ sharks
$$z =\frac{ 21.5 - 20 }{ \frac{1.209 }{ \sqrt 4}}$$
I came up with $2.168870...$ for the test statistic.
For critical value of $z$, I used the given alpha to find the value of $2.326$ from a confidence interval of 98%.
What are the answers and what am I doing wrong ?
(Thank you for the edit)

@Flounderer's Answer is correct (+1). This is for clarification, intuition, and
verification. 
You cannot to a z-test here. The population standard deviation $\sigma$
is not known, and so has to be estimated by the sample standard deviation
$S = 1.291.$ So your test statistic is
$$T = \frac{\bar X - \mu_0}{S/\sqrt{n}}.$$
You should not write $\sigma$ in the denominator because you do not
know it. 
The critical value for t in this one sided test cuts off 1% from the upper
tail of Student's t distribution with 3 degrees of freedom. So the
critical value is 4.541. You should look in row 3 of the printed t table
in your text to find that value.

Here is a printout of this one-sided one-sample t test from Minitab 17
statistical software.
One-Sample T: Length 

Test of μ = 20 vs > 20

Variable  N    Mean  StDev  SE Mean     T      P
Length    4  21.500  1.291    0.645  2.32  0.051

The P-value 0.051 indicates that this null hypothesis could
not (quite) be rejected at the 5% level because the P-value is
slightly above 0.050. But you want to test at the 1% level
and the decision not to reject is nowhere near the borderline.
In terms of the critical value 4.541 from the t table, you cannot
reject because the computed value of the T statistic is 2.32.
You could reject at the 1% level only if $T > 4.541.$

Just based on tuition, you should not expect to reject $H_0: \mu \le 20$
against the alternative $H_a: \mu > 20.$
One of the sharks captured is 20 ft long and the other three are only
a little longer. Expensive and dangerous or not, catching just four
sharks is not enough.
According to a more advanced computation, if the true length of sharks off the Bermuda were 22 ft (with a SD around 1.3), then a sample of size of at least $n = 6$ would have been required in order to have a reasonable chance of detecting that they average over 20 ft.
