Solution:

The answer is $\mathbf{( B )}$. First observe that (for $n \geq 2$ ) the relation $F_{n+1}=F_{n}+F_{n-1}$, together with the fact that $F_{n-1} \leq F_{n}$, implies that $F_{n}$ is at least half of $F_{n+1}$. Now let $k \geq 1$ and let $F_{n}$ be the smallest Fibonacci number with $k+1$ decimal digits (hence $F_{n} \geq 10^{k}$ ). We have $F_{n-1} \geq \frac{1}{2} \cdot 10^{k}$ and therefore $F_{n+1}=F_{n}+F_{n-1} \geq\left(1+\frac{1}{2}\right) \cdot 10^{k}$. Proceeding in the same way we successively obtain the inequalities $F_{n+2} \geq\left(\frac{3}{2}+1\right) \cdot 10^{k}, F_{n+3} \geq\left(\frac{5}{2}+\frac{3}{2}\right) 10^{k}, F_{n+4} \geq\left(4+\frac{5}{2}\right) \cdot 10^{k}$ and $F_{n+5} \geq\left(\frac{13}{2}+4\right) \cdot 10^{k}>10^{k+1}$, from which we see that $F_{n+5}$ has at least $k+2$ decimal digits.

On the other hand, by definition we have $F_{n-2} \leq F_{n-1}<10^{k}$, from which we successively obtain $F_{n}<2 \cdot 10^{k}, F_{n+1}<3 \cdot 10^{k}, F_{n+2}<5 \cdot 10^{k}, F_{n+3}<8 \cdot 10^{k}$, so $F_{n+3}$ still has $k+1$ decimal digits. In conclusion, given any number $k \geq 2$, there are at least 4 and at most 5 Fibonacci numbers with $k$ digits.