"Must be an integer" means the lowest possible value of $y$.

$450y=n^3$ --> $2*3^2*5^2*y=n^3$. As $n$ and $y$ are integers, $y$ must complete the powers of 2, 3, and 5 to cubes (generally to the multiple of 3).

Thus $y_{min}=2^2*3*5$, in this case $2*3^2*5^2*y=(2*3*5)^3=n^3$.

Notice that for this value of $y$ only the first option is an integer: $\frac{y}{{3*2^2*5}}=\frac{2^2*3*5}{{3*2^2*5}}=1$.

Answer: B.