Given the half-life, you have enough information to find the decay factor \(k\) that satisfies
\(\dfrac12=e^{55k}\implies k=-\dfrac{\ln2}{55}\approx-0.013\)
Starting with 1 kg of the isotope, after 120 days you would be left with
\(y=e^{120k}=\dfrac1{2^{24/11}}\approx0.220\text{ kg}\)
After 240 days, you would have
\(y=e^{240k}=\dfrac1{2^{48/11}}\approx0.049\text{ kg}\)
