For 6A to be a perfect cube, then you must have (2*3)(A) = n^2, for some number n. This number n must include the factors 2 and 5, along with whatever A brings to the table. For 2*3*A to be a cube, then there must be two 2s and two 3s as factors of A. So A = (2^2)(3^2)(something else cubed), so n^3 = (2^3)(3^3)(something else cubed).
So what factors, at a minimum, must be included in A? (Note that the "something else squared" and "something else cubed" could be just 1^2 and 1^3, respectively.) For instance, since you [i]must have a 5 in A (from the squaring) and 6A must be a cube, how many copies of 5 must be in A?
If you get stuck, please reply showing your work and reasoning so far. Thank you!