Answer:
The minimum sample size required for the estimate is 345.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.
The standard deviation is known to be 1.8.
This means that 
What is the minimum sample size required for the estimate?
This is n for which M = 0.19. So






Rounding up to the next integer:
The minimum sample size required for the estimate is 345.