Suppose that a particular human trait is caused by a recessive x-linked allele. The frequency of this recessive allele is 0.02 in a given population. Assume that the dominant and recessive alleles are in Hardy–Weinberg equilibrium and that the population consists of an equal number of males and females. What percent of the population will exhibit the recessive trait?

Hardy–Weinberg equilibrium is a formula linking frequencies of genetic traits to frequencies of affected individuals within a population. The general formula is
p^(2)+2pq+q^(2)=1
where "p" is the frequency (between 0 and 1) of the dominant allele in the population, and "q" is the frequency (between 0 and 1) of the recessive allele in the population. In order for this equation to remain valid, there must only be two alleles in the population. This is mathematically stated as p + q = 1.
In this equation, p² represents the proportion of people who are homozygous for the dominant allele, 2pq represents the proportion of people who are heterozygotes, and q² represents the people who carry the recessive trait.
In this answer, we are told the frequency of the recessive allele is 0.02. This means the frequency of the dominant allele is 0.98 (or 98%).
For the rest of this problem, I will refer to Xp and Xq as the dominant and recessive alleles, respectively.

The full Hardy–Weinberg equilibrium equation is applicable to females, since they have two X-chromosomes. In this case, the frequency of females that will exhibit the trait is q², or .0004 (0.04%).

Since males only have one X-chromosome, the frequency of affected males will be equal to the frequency of the recessive trait in the population. In this case, this frequency of affected males is 0.02, or 2%.

To combine affected female frequency and male frequency to calculate a population frequency, we multiply each population by their fraction of the total population. In other words, (frequency affected females)*(fraction of population that is female) + (frequency affected males)*(fraction of population that is male) is the total population frequency of affected individuals. For this problem, this becomes (0.0004)*(0.5)+(0.02)*(0.5) = .0102, or 1.02%.

The answer to your question is that 1.02% of the population will exhibit an X-linked recessive characteristic when the frequency of the recessive X-allele is 0.02.