Answer: D. $100(b-1)$

Explanation:1. Evaluate $f(x)$ and $f(x+1)$: $f(x) = 66b^x$ When x increases by $1$, the new value is $x+1$. $f(x+1) = 66b^{x+1} = 66b^x \cdot b^1 = (66b^x) \cdot b = f(x) \cdot b$

2. Understand "increases by c%": If a value $V$ increases by c%, the new value $V\_{new}$ is $V + \dfrac{c}{100}V = V(1 + \dfrac{c}{100})$. So, $f(x+1) = f(x) (1 + \dfrac{c}{100})$.

3. Equate the two expressions for $f(x+1)$: From step $1$, $f(x+1) = f(x) \cdot b$. From step $2$, $f(x+1) = f(x) (1 + \dfrac{c}{100})$. Therefore, $f(x) \cdot b = f(x) (1 + \dfrac{c}{100})$.

4. Solve for b in terms of c (or vice versa): Since $f(x) = 66b^x$ and b is positive, $f(x)$ is not zero. We can divide both sides by $f(x)$: $b = 1 + \dfrac{c}{100}$

5. Solve for c in terms of b: We need to find an expression for c. $b - 1 = \dfrac{c}{100}$ Multiply both sides by $100$: $100(b - 1) = c$ So, $c = 100(b-1)$.

6. Check the constraints: Given $0 < c < 100$. Since $c = 100(b-1)$, then $0 < 100(b-1) < 100$. Dividing by $100$: $0 < b-1 < 1$. Adding $1$ to all parts: $1 < b < 2$. The problem states b is a positive constant. If there's an increase (c>0), then $b-1 > 0 \Rightarrow b > 1$. If the increase is less than $100\%$ ($c<100$), then $b-1 < 1 \Rightarrow b < 2$. So, the condition $1 < b < 2$ is consistent with an increase between $0\%$ and $100\%$.

7. Check the solution with the options: The expression for c is $100(b-1)$. This matches option [D].

8. Conclusion: The expression that gives the value of c in terms of b is $100(b-1)$.