This is pretty similar to a famous exercise about the rate of convergence of the sequence defined by $a_{n+1}=\sin(a_n)$. In your case it is pretty simple to notice that $\{a_n\}$ is positive and decreasing, hence convergent to zero, and by setting $b_n=1/a_n$ (as you cleverly did) we have
$$ b_{n+1} = b_n + \frac{b_n}{3b_n-1}= b_n+\frac{1}{3}+\frac{1}{9b_n-3}\tag{1}$$
where
$$ b_{n+1} \geq b_n + \frac{1}{3}$$
implies
$$ b_n \geq \frac{n+2}{3} \tag{2}$$
and $a_{100}\leq \frac{3}{102}$, which only leaves options $(A)$ and $(B)$. At this point one might guess that $(2)$ is pretty tight, hence $100 a_{100}$ is less than $3$ but pretty close to $3$ (option $\mathbf{(B)}$). Proving this requires a lower bound for $a_n$, i.e. an upper bound for $b_n$, which we can obtain by plugging $(2)$ back into $(1)$ (bootstrapping):
$$ b_{n+1} \leq b_n + \frac{1}{3} + \frac{1}{3}\cdot \frac{1}{n+1} \tag{3}$$
leads to
$$ b_n \leq \frac{n+H_n+1}{3} \tag{4}$$
hence $100 a_{100}\geq \frac{300}{101+H_{100}}\geq 2.8 $ and $\mathbf{(B)}$ actually is the correct answer.
An accurate approximation of $H_n=\sum_{k=1}^{n}\frac{1}{k}$ comes from $H_n\approx \log(n)+\frac{1}{\sqrt{3}}$.
A more elementary approximation of $H_{100}$ (a worse one, but still effective for our purposes) can be derived without calculators, using just telescopic series and the Cauchy-Schwarz inequality:
$$\begin{eqnarray*} H_{100} = \frac{25}{12}+\sum_{k=5}^{100}\frac{1}{k} \leq \frac{25}{12}+\sum_{k=5}^{100}\frac{1}{\sqrt{k(k-1)}}\leq \frac{25}{12}+\sqrt{\sum_{k=5}^{100}1 \sum_{k=5}^{100}\frac{1}{k(k-1)}}\end{eqnarray*}$$
leads to
$$ H_{100} \leq \frac{25}{12}+\sqrt{96\sum_{k=5}^{100}\left(\frac{1}{k-1}-\frac{1}{k}\right)}=\frac{25}{12}+\sqrt{96\left(\frac{1}{4}-\frac{1}{100}\right)}\leq \frac{25}{12}+\sqrt{24} \leq 7 $$
then to $100\,a_{100} \geq 3-\frac{1}{4}$.
Best Answer
Hint:
$$\frac12+\frac14+\ldots+\frac1{200} =\frac12\left(1+\frac12+\ldots+\frac1{100}\right)$$
and you may want to add it twice to both sides