smc

Note $R_n={\sum }_{k=0}^{n-1}10^k=\frac{10^n-1}{10-1}$. Therefore $\frac{R_{24}}{R_4}=\frac{10^{24}-1}{10^4-1}$. We can recognize this is also the formula for the sum of a geometric series $1+10^4+(10^4{)}^2+\cdots +(10^4{)}^5=1+10^4+10^8+\cdots +10^{20}$ Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the $10^5$, $10^6$ and $10^7$ places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives $5\times 3=15$ zeros altogether. The answer is $\boxed{15}$