From the sample population of 100 people, the proportion who had kids is 75/100=¾ or 75%.
We're not given the sample standard deviation (SSD) so we need to calculate it. If 75% had kids, then 25% did not have kids. The variance is 100(0.75)(0.25)=18.75, so the SSD is √18.75=4.33% approx. For the sample, we can say that the proportion is 75%±4.33%, or 0.75±0.0433. But this SSD applies to the sample, not the whole population, so we have to make a standard adjustment for the sample: 4.33/√100=0.433% approx.
To bring in the 95% confidence interval (CI) we work on the Z score for the normal distribution. Z=|X-75|/0.433 where X% is a measure of the proportion in lots of different samples. The CI tells us how far X can be on either side of the mean proportion of 75% so as to lie within 95% of the results of measuring many samples. The significance level for a CI of 95% is (100-95)/2=2.5% because we have to remember that the normal distribution is symmetrical about the mean proportion, so that 5% is distributed equally on both sides of the mean in the tails of the distribution. The Z value is about ±1.96 for 95%, so |X-75|/0.433=1.96, |X-75|=0.849, X=75.849 or 74.151.
Therefore 74.151%<p<75.849%, that is, the proportion can be expected to lie between 74.151% and 75.849% in 95% of the cases.
In decimals this is 0.742<p<0.758 (3 decimal places).