There's a much simpler way:
The probability of all students being able to solve the problem is 0.84×0.7×0.6=0.3528.
Therefore 1-0.3528=0.6472 is the probability that at least one student wasn’t able to solve the problem.
Since you came to the same conclusion, you're correct if I'm correct.
HOWEVER:
The question states that the problem was solved and we know that the probability of no-one solving it is 0.16×0.3×0.4=0.0192. The probability of someone solving it is therefore 1-0.0192=0.9808. The probability of 0.6472 includes the outcome that no-one solved the problem, and this has to be discounted. Doesn't that mean that we need to deduct 0.0192 from 0.6472=0.628?
SUMMARY
Consider all possible outcomes:
(1) All 3 students solved the problem;
(2) Some students solved the problem, some didn't; (in this question, some means one or two)
(3) None of the students solved the problem.
The sum total of these probabilities=1. When we exclude (1) and (3) we get what was asked for, so we just have to calculate probabilities for (1) + (3) and subtract from 1. 1-0.3528-0.0192=0.628.