<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:

where,
= half life of the reaction = 5.26 years
Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = ? yr
= initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 75) = 25 grams
Putting values in above equation, we get:

The original sample was purchased in June 2016
As, June is the 6th month of the year, which means the time period will be 
Adding the time in the original time period, we get:

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027