I am so stuck here, can you please review and tell me where I went wrong?

Annual starting salaries for college graduates with business
administration degrees are believed to have a standard deviation between 30,000 and 45,000. Assume that a 95% confidence interval estimate of the mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is?
A: $500 =(.0167)(.9833)(1.96/.025)^2 = 101 (.0111)(.99)(1.96/.025)^2= 66
B: $200 = (.0067)(.99) (1.96/.025)^2 = 41 (.0044)(.99) (1.96/.025)^2 = 27
C: $100 = (.0033)(.99) (1.96/.025)^2 = 20 (.0022)(.99) (1.96/.025)^2 = 17

A. A sample size between 101 and 66 is needed for a desired
margin of $500, assuming a 95% confidence interval.
B. A sample size between 41 and 27 is needed for a desired
margin of $200, assuming a 95% confidence interval.
C. A sample size between 20 and 17 is needed for a desired
margin of $100, assuming a 95% confidence interval.