A vial of iodine-131 has an activity of 640 MBq on Monday…

Step 1. After $28$ days. The number of half-lives is $\dfrac{28}{8.0} = 3.5$. So $R = 640\ \text{MBq} \times 2^{-3.5} = 640\ \text{MBq} \times 0.0884 = 56.6$ MBq.

Step 2. Down to $20$ MBq. $\dfrac{640}{20} = 32 = 2^5$, so $5$ half-lives are needed. $t = 5 \times 8.0\ \text{d} = 40$ days.

Step 3. The first $4.0$ days. That is $0.5$ half-life. The fraction left is $2^{-0.5} = 0.707$, so $1 - 0.707 = 0.293$ of the nuclei have decayed, which is $29.3\ \%$.