I don't know about any internet links but I can definitely help you solve this question:

n=(k+1)^3 and we know k>1.

Put values of K and see what comes out as the reminder of n/k.

Take k=2, => n=3^3 = 27. n/k= 27/2. This will leave a reminder of 1.
Take k=3, => n=4^3 = 64. n/k = 64/3. This will again give reminder 1. (I hope you get how the reminder comes out to be 1).

Take k=4, n=5^3 = 125. again, n/k =125/4 leaving a reminder of 1.

So, in this way, if we take any valve of k, we will always get the reminder 1. Option 1 is sufficient to answer the question.

Option 2 (k=5) is not sufficient because it says nothing about n.

Another alternate method which is actually easier would be:

n=(k+1)^3, expanding this we get, n= k^3 + 3k^2 + 3k + 1.--- (1)

Now, to get n/k, we simply divide both sides of (1) by k.

We get, n/k = k^2 + 3k + 3 + 1/k.

So, it is clear that since k>1, whatever the value of k, the reminder of n/k will always be 1.

Again, 1 is sufficient.