Reducing the Proportion of Phenotype in Hardy Weinberg Population through Selection
The theory

You have to use the diploid selection equation (see here). Let $p$ be the the frequency of the recessive allele of interest, you get

$$p' = \frac{p^2 W_{aa} + p(1-p)W_{aA}}{\bar W}$$

, where

$$\bar W = p^2 W_{aa} + 2p(1-p)W_{aA} + (1-p)^2W_{AA}$$

You can simply iterate over with this equation. Do no forget to recompute $\bar W$ at each generation as the allele frequency changes, so does the mean fitness $\bar W$. Also, do not forget that we are looking for the generation where the frequency of sick individuals is lower than $0.05$, that is the generation where $p&lt;\sqrt{0.05}$.

I don't think this recursion equation has any general solution unfortunately. This would mean that iteration is the only way to got.

The practice

I am a bit too lazy to compute everything by hand so here a short R code that will do the calculations for us.

frequencySickIndividuals = 0.16
p = sqrt(frequencyHomozygoteRecessiveMutants)
waa = 0.5
waA = 1
wAA = 1
for (generation in 0:10)
{
    frequencySickIndividuals = p^2
    cat(paste0("Generation ", generation, " frequency sick individuals = ", frequencySickIndividuals, "\n"))
    p = (p^2 * waa + p*(1-p)*waA) / (p^2*waa + 2*p*(1-p)*waA + (1-p)^2*wAA)
}


It outputs

Generation 0 frequency sick individuals = 0.16
Generation 1 frequency sick individuals = 0.120982986767486
Generation 2 frequency sick individuals = 0.0935350533786218
Generation 3 frequency sick individuals = 0.0738632061510922
Generation 4 frequency sick individuals = 0.0594638505565122
Generation 5 frequency sick individuals = 0.0487003101360286
Generation 6 frequency sick individuals = 0.0404940467744661
Generation 7 frequency sick individuals = 0.034123112395776
Generation 8 frequency sick individuals = 0.0290951856295452
Generation 9 frequency sick individuals = 0.0250680151553009
Generation 10 frequency sick individuals = 0.0217991643532054


So, it actually takes 5 generations to reach an allele frequency lower than $\sqrt{0.05}$. 

Source of information on hardy-Weinberg problems

Note by the way, you might be interested in having a look at a tutorial on how to solve Hardy-Weinberg problems on the post Solving Hardy Weinberg problems.