Answer: C

Small Hint:

Write the rational common ratio in lowest terms as $\frac{c}{d}$

Big Hint:

Integrality forces the middle term to be a multiple of $c^2d^2$; use that $211$ is prime

Solution:

Let the common ratio be $\frac{c}{d}$ in lowest terms and the middle term be $a.$ Since all five terms are integers, $a$ is divisible by $c^2d^2,$ so write $a=kc^2d^2.$ The sum $211$ is then divisible by $k.$ Thus $k$ is $1$ or $211.$ But the middle term $a$ is less than $100,$ so $k=1.$

Thus $$\begin{aligned} d^4+d^3c+d^2c^2&\\ \quad+dc^3+c^4&=211. \end{aligned}$$ The bound gives $c,d\lt4.$ If either is $1,$ the possible sums are $5,$ $31,$ or $121,$ not $211.$ Coprimality therefore leaves $c$ and $d$ equal to $2$ and $3$ in either order. The progression is $16,$ $24,$ $36,$ $54,$ $81.$ Its square terms sum to $16+36+81=133.$

Therefore, the correct answer is C.