Solution:

We need to find the smallest possible $n$ such that $$1- \dfrac n{100} < (0.85)^2,$$ $$1- \dfrac n{100} < (0.9)^3,$$ $$1- \dfrac n{100} < (0.75)(0.95).$$ Note that $$0.75\cdot 0.95 = 0.85^2-0.1^2 < 0.85^2,$$ so we don't need to worry about the first condintion since the last condition is true. Then, the second condition yields $$1- \dfrac n{100} < 0.729.$$ $$n > 27.1.$$

Also, we also can see that $$1- \dfrac n{100} < 0.95\cdot 0.75$$$$1- \dfrac n{100} < \dfrac 34 \cdot \dfrac{19}{20}$$$$1- \dfrac n{100} < \dfrac{57}{80}$$$$1- \dfrac n{100} < \dfrac{285}{400}.$$

Therefore, $$n > \dfrac{115}{4} = 28.75.$$ Combining our conditions yields a smallest $n$ of $n=29.$

Thus, the correct answer is C .