A random sample of $30$ days gives a mean distance of $350$ km, where the population standard deviation is known to be $60$ km. Which is the $98\%$ confidence interval for the mean distance travelled, to 1 decimal place?

Because the population standard deviation is known, a $z$ interval is used. The standard error is $\dfrac{60}{\sqrt{30}} = 10.9545$, and for $98\%$ confidence $z = 2.326$, so the margin of error is $2.326 \times 10.9545 = 25.48$. The interval is $350 \pm 25.48 = (324.5,\ 375.5)$. Dividing by $\sqrt{n}$ is the step that turns the spread of single days into the spread of the sample mean; leaving it out gives the far wider first option.