Answer
1.78 billion years.
Work Step by Step
Assume that $^{87}_{38}Sr$ is stable (it’s not listed in Appendix B). We are told that initially, there was none of it in the rock. Thus every atom of Rb that decayed is now Sr-87.
$$N_{Sr}=-\Delta N_{Rb}$$
Assume that the decay rate is constant (which is to say that the elapsed time is much smaller than the half-life of Rb). Use equation 30-3a.
$$-\Delta N_{Rb}=\lambda N_{Rb}\Delta t$$
Solve for the elapsed time by combining the previous relationships.
$$\Delta t=\frac{N_{Sr}}{ N_{Rb}}\frac{T_{1/2}}{ln 2}$$
$$\Delta t=(0.0260)\frac{4.75\times10^{10}y}{ln 2}=1.78\times10^9 y$$
The age of the rock is less than 4 percent of Rb-87's half-life, so it was all right to assume that the decay rate was constant over the 1.78 billion years.
