Hello Vjesus12.

Let's see your question.

You could try number by number or solve it as follows:

$$\sqrt{n\cdot7!}=\sqrt{n\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=2\sqrt{n\cdot7\cdot(3\cdot2)\cdot5\cdot3\cdot2}$$ $$2\sqrt{n\cdot7\cdot2\cdot3\cdot5\cdot3\cdot2}=2\sqrt{n\cdot7\cdot4\cdot9\cdot5}=2\cdot2\cdot3\sqrt{n\cdot7\cdot5}=12\sqrt{n\cdot7\cdot5}.$$ Now, this last number is an interger when the prime factorizazion has an odd number of 5's and a 7's.

On the given list, the smallest number that holds this condition is 35.

So, the correct answer is [spoiler]B=35[/spoiler].

I hope this explanation may help you.

I'm available if you'd like a follow-up.

Regards.