Question 678364
STARTING NOTES:
I would expect that by "30 ounces" of solution the problem means 30 fluid ounces, a measure of volume. It would be weird if it meant that the solution weighed 30 avoidupois ounces (1 pound and 14 ounces).
It makes me believe we are measuring the volume of water to be added in fuid ounces.
It also suggests that the solution is a 6% w/v (weight in volume) solution, containing 6 grams of salt per 100 milliliters of solution. 
 
THE PROBLEM
What follows may not be how your teacher expects you to solve the problem, because it is done differently in different classes and books.
On diluting, the concentration is reduced by the dilution factor, so if you double the initial volume, you halve the concentration; if you triple the initial volume, the final concentration is one third of the initial concentration, and so on.
The dilution factor is the ratio of volumes, (final volume)/(initial volume).
The dilution factor is the same as the ratio of concentrations, (initial concentration)/(final cocentration).
To get the concentration from 6% to 1.8%, the dilution factor needed is
{{{6/1.8=60/18=10/3=100/30}}}.
(Those ratios are the same. It's the same proportion)
So you need to go from 30 ounces to a final volume of 100 ounces.
In math problems, as you dilute a solution volumes are additive, so if {{{x}}} ounces of water are added to 30 ounces of solution, the total volume of the diluted solution would be {{{30+x=100}}} ounces.
{{{30+x=100}}} --> {{{30+x-30=100-30}}} --> {{{highlight(x=70)}}}
so {{{highlight(70)}}} ounces of water should be added.
 
THE SCIENCE RANT:
The name ounce is applied to differenty units of force, mass (avoirdupois ounce, troy/apothecary ounce) and volume (fluid ounce), and makes for some confusion.
Scientists do not use ounces, and specify when in doubt. (If a scientist says a metric ton, I know it is 1000 kilograms, but an Iowa farmer who says a ton probably means 2000 pounds, which is not the same).
As an analytical chemist, I would probably express the concentration of a salt solution as 6% w/v (weight in volume) meaning 6 gram salt in 100 milliliters of solution.
On rare occasions, I would be giving a concentration in terms of weight in weight, as in 6% w/w, meaning 6 grams of salt in 100 g of solution, which would have a volume of about 96 milliliters, because the density would be 1.04 gram per milliliter. So 6.0% w/w =6.2% w/v.
Volumes are not exactly additive in the real world, although in may cases they are almost exactly additive.
The density of a salt solution changes a bit with concentration, and also with temperature. At {{{20^o}}} Celsius, density is 1.04 g/mL for salt solutions at concentrations of about 6.0% and 6.2% w/v. It is 1.01 g/mL for 1.8% w/v salt solutions.
In the case of this problem, adding 70.0 fluid ounces of water to 30.0 fluid ounces of 6.000% w/v solution would yield 100.2 fluid ounces of a solution that would not be exactly 1.800% w/v, but the difference is too small to matter to a nurse diluting a solution.