Explanation

(c) The activity $\left( - \frac{dN}{dt} \right) = \lambda N \Rightarrow N = \left( - \frac{dN}{dt} \right)\left( \frac{T_{1/2}}{\log_{e}2} \right)$
Taking the ratio of this expression for $\ ^{240}Pu$ to this same expression for $\ ^{243}Am$
$\frac{N_{Pu}}{N_{Am}} = \frac{\left( - \frac{dN_{Pu}}{dt} \right)\left( T_{1/2} \right)_{Pu}}{\left( - \frac{dN_{Am}}{dt} \right)\left( T_{1/2} \right)_{Am}} = \frac{(5\mu ci) \times (6560y)}{(4.45\mu ci) \times (7370y)} = 1$
$i.e.,$ the two samples contains equal number of nuclei