Rearrange the equation: 1.96(34.15/√n)=13.03-12.98=0.05,
34.15/√n=0.05/1.96,
√n=34.15/(0.05/1.96)=34.15(1.96)/0.05=66.934/0.05=1338.68,
n=1,792,064 approx. This answer is suspiciously large, and 34.15 is rather large for a standard deviation with an apparent mean of only 13.03. Are you sure you have the right SD? SD of 3.415 would be more realistic. √n=133.868, n=17,920. X=μ±1.96s/√n, where μ is the mean=13.03 and s the sample SD. The low end of the 95% CI would then be X=μ-1.96s/√n. The value 12.98 would then be comfortably within the CI.
The population SD and the sample SD can be regarded to be the same if the sample size is sufficiently large, that is, comparable to the population size.