Answer
The fraction of plutonium that is left is so small that it is essentially zero. We can assume that all of the original sample of plutonium would decay after this length of time.
Work Step by Step
When half of a sample of $^{238}U$ has decayed, the time that has passed is $4.5\times 10^9~y$
We can express this time period as the number of half-lives of curium:
$\frac{4.5\times 10^9~y}{3.4\times 10^5~y} = 13,235$
We can find the fraction of curium that is left:
$(\frac{1}{2})^{13,235} \approx 0$
The fraction of plutonium that is left is so small that it is essentially zero. We can assume that all of the original sample of plutonium would decay after this length of time.
