The critical chi^2 value of 79.08 (in this case) would need to be provided, likely as a value in the overall lookup table. Excel function = CHISQ.INV.RT(5%, 60) = 79.08 verifies it. I don't think the question is especially well-written, it assume a bit too much familiarity in my opinion. As usual the null can be either one- or two-tailed; this question assumes a one-tailed test (which is the weakness: a question should not use the answers to convey the one-tailed assumption. Here, you have to infer a one-tailed test by noticing that all answers reflect that assumption).
So, the 79.08 value corresponds to the critical value such that 5% of the chi-square distribution (the "reject" region) is entirely in the right-tail (which, in turn, counterintuitvely, corresponds to a lower bound on volatility). In other words, around the sample volatility of 21.0%, a two-tailed 90% confidence region, which is informed by the critical value (in term which depends on df), will be bounded by 18.3% (lower) and 24.75% (upper). The 18.3% lower two-tailed corresponds to a 95% one-tailed, such that anything less than 18.3% falls into the 95% one-tailed reject region; I got 18.3% with 18.3% = sqrt(60*21%^2/79.08).
IMO, the hard part here is the one vs two tail, and as I mentioned, this question simply assumes one-tailed test. Given, or having looked up, the critical value of 79.08, it corresponds either to a one-tailed 95% test, or to one side of a two-tailed 90% test.
As this can be confusing, let me put it one more way: the critical chi^2 of 79.08 is analogous to the critical Z value of 1.645, as both are quantiles which identify the "beginning" of the 5% right-tail rejection region, such that test values greater than them will reject a one-sided 95% confidence test (although the chi-square is not symmetrical and morphs with df!).