Well, since the ordering relation is $|$ (divides), then, you can rephrase the "least upper bound" question as "what number, is divisible by 6, 10, 14, and 15, and such that no divisor of it is itself divisble by 6, 10, 14, and 15". Now, "is divisible by x" is the same as "is a multiple of x", so we see that we want a multiple of 6, 10, 14, and 15 that is not itself a multiple of such a multiple. A candidate is the least common multiple (or lcm) of 6, 10, 14, and 15. This will be divisible by all of those, and furthermore, we can prove it is the desired least upper bound since it is not divisible by another such multiple, since such a multiple would have to be a smaller one, but it's the least one (note that if $n | m$, then $n \le m$. Since $n \ne m$ in this case, $n < m$). We have $\mathrm{lcm}(6, 10, 14, 15) = 210$, so 210 is the least upper bound.
Now, for the greatest lower bound. We can rephrase this as "what number is a divisor of 6, 10, 14, and 15, and such that it is not itself a divisor of such a divisor". A natural candidate is then the greatest common divisor (or gcd) of 6, 10, 14, and 15. This will be a divisor of all of those. To prove this is the desired number, we merely note that if it was a divisor of another such divisor, then there would be a larger common divisor, but it's the greatest one. We have $\gcd(6, 10, 14, 15) = 1$. BUT!!! $1$ is not in the set! So, there is NO greatest lower bound. There are in fact no lower bounds at all, since there are no common divisors except $1$!