Using Amdahl's law how do you determine execution time after an improvement?
Since the FP speedup  slows down  the cache by a factor of $2/3$, the cache is "sped up" by $3/2$. Hence the overall speedup is

$ S = \frac{1}{0.7 + 0.2/2 + 0.1 \times \frac {3}{2}} = 1.05 \equiv 5\%$

The execution times now have the ratio: $0.7:0.1:0.15$ for remaining operations:FP:cache. The percentage time spent on FP now is $\frac{0.1}{0.7+0.1+0.15} = 0.105 \equiv 10.5\%$.