This question is equivalent to asking:

Consider a program in which 90 % of the of the part can be sped up and the speed up is 30 times. What is the overall speedup?

Simple application of Amdahl's law, the answer should be $\frac{1}{0.1 + \frac{0.9}{30}}$

However, if you want to see it as using the memory itself, let $t_a$ be the time to access the main memory, and let $t_b$ be the time used to access the cache.

Then, $$\text{Speedup} = \frac{t_a}{h \cdot t_b + (1-h) \cdot t_a}$$

Here, $\frac{t_a}{t_b} = 30$, so we essentiallly get the same result.