Q.What is the amount of $_{27}^{60}$Co necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?

[!TLDR] $\lambda=0.693/5.3$ yr $=4.144\times10^{-9}$ s$^{-1}$; $A=10.0$ mCi $=3.7\times10^8$ s$^{-1}$; $N=A/\lambda\approx8.93\times10^{16}$; mass $=N\times60/N_A\a …

Convert the half-life to a decay constant: $\lambda=\frac{0.693}{5.3\text{ yr}}=0.1307$ yr$^{-1}=\frac{0.1307}{3.156\times10^7\text{ s/yr}}\approx4.144\times10^{-9}$ s$^{-1}$. Convert the required activity to becquerel: $10.0\text{ mCi}=10.0\times10^{-3}\times3.7\times10^{10}=3.7\times10^8$ decays/s. The number of Co-60 nuclei needed is $N=A/\lambda=\frac{3.7\times10^8}{4.144\times10^{-9}}\approx8.928\times10^{16}$. Converting to mass using the isotope's molar mass (approximately 60 g/mol) and Avogadr …

Find $\lambda$ from the half-life, convert the required activity from mCi to Bq, solve $N=A/\lambd …

Leaving the given activity in millicurie without converting to becquerel (decays per second) before dividing by $\lambda$ (which is in SI, per-second …