Anurse at a local hospital is interested in estimating the birth weight of infants. how large a sample must she select if she desires to be 98% confident that the tue mean is within 3 ounces of sample mean. the standard deviation of birth weights is known to be 6 ounces

For large sample confidence intervals about the mean you have: 

xBar ± z * sx / sqrt(n) 

where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
 
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.

z * sx / sqrt(n) = width. 

so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348 

The equation we need to solve is:
z * sx / sqrt(n) = width 
n = (z * sx / width) ^ 2. 
n = ( 2.326348 * 6 / 3 ) ^ 2 
n = 21.64758 

Since n must be integer valued we need to take the ceiling of this solution.
n = 22 

The answer is 76° i believe