Answer:
The smallest sample size required to obtain the desired margin of error is 44.
Step-by-step explanation:
I think there was a small typing error, we have that  is the standard deviation of these weighs.
 is the standard deviation of these weighs.
We have that to find our  level, that is the subtraction of 1 by the confidence interval divided by 2. So:
 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of  .
.
So it is z with a pvalue of  , so
, so 
Now, find the margin of error M as such

 Which of these is the smallest approximate sample size required to obtain the desired margin of error?
This sample size is n.
n is found when 
So



Simplifying by 15



Rounding up
The smallest sample size required to obtain the desired margin of error is 44.