Explanation

(c) $\lambda_{\alpha} = \frac{1}{1620}per\ year$ and $\lambda_{\beta} = \frac{1}{405}per\ year$ and it is given that the fraction of the remained activity $\frac{A}{A_{0}} = \frac{1}{4}$
Total decay constant
$\lambda = \lambda_{\alpha} + \lambda_{\beta} = \frac{1}{1620} + \frac{1}{405} = \frac{1}{324}per\ year$
We know that $A = A_{0}e^{- \lambda t} \Rightarrow t = \frac{1}{\lambda}\log_{e}\frac{A_{0}}{A}$
$\Rightarrow t = \frac{1}{\lambda}\log_{e}4 = \frac{2}{\lambda}\log_{e}2 = 324 \times 2 \times 0.693 = 449\ years$