Answer: B

Small Hint:

Write the populations as $a^2,$ $b^2+1,$ and $c^2$

Big Hint:

From $b^2-a^2=99,$ test the positive factor pairs of $99$, then impose $c^2-a^2=200$

Solution:

Let the original population be $a^2.$ Then $$\begin{aligned} b^2-a^2&=99,\\ c^2-a^2&=200. \end{aligned}$$ From $(b-a)(b+a)=99,$ the positive possibilities for $a$ are $49,$ $15,$ and $1.$ Checking the second condition, only $a=49$ works, since $$49^2+200=2601=51^2.$$ The population is $49^2=7^4,$ a multiple of $7.$

Thus, the correct answer is B.