Answer
It takes about $672$ years for a given amount of this material to be reduced by one quarter.
Work Step by Step
Using Exercise 13, we know that the decay of Radium-226 is described by the differential equation $Q' = -rQ$, with general solution $Q = Q(t) = ce^{-rt}$, where $c$ is a constant depending on the initial amount of Radium-226, $r$ is its decay-rate, $t$ is the number of years.
Using Exercise 13 again, if we denote $\tau = 1620$ years the half-life of Radium-226, we know that it satisfies the equation $r \tau = ln(2) \iff r = ln(2)/\tau = ln(2)/1620$.
Now, if we want to know the numbers of years $t^*$ it takes so that the initial amount of material is reduced by $1/4$, we get the equation :
$$Q(t^*) = Q(0) - 1/4Q(0) = 3/4Q(0)$$
or equivalently
$$ce^{-rt^*} = 3/4ce^{-0r} = 3/4c \iff e^{-rt^*} = 3/4.$$
By applying the natural logarithm to both sides of the equation, we get $-rt^* = ln(3/4) \iff t^* = ln(3/4)/(-r) = -[1620\cdot ln(3/4)]/ln(2) \approx 672$ years.
