Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 39

$3.90\times 10^3y$

We know that $\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{432y}=0.00160y^{-1}$ Now, $R=\frac{R_{\circ}}{525}$ We also know that $t=\frac{1}{\lambda}\ln \frac{R_{\circ}}{R}$ $\implies t=\frac{432 y}{\ln 2}(\ln \frac{R_{\circ}}{\frac{R_{\circ}}{525}})$ $t=3.90\times 10^3y$