Answer
$65.396$ amu
Work Step by Step
We simply take a weighted average:
$$\sum_{n=1}^{N} w_{n}m_{n} = M$$
where $w_{n}$ is the atomic average and $m_{n}$ is the atomic weight of a specific zinc isotope, $n$. $M$ is the final average weight that we wish to calculate. $N$ is the total number of isotopes; in this case, it is 5
$.4863(63.929)+.2790(65.926)+.0410(66.927)+.1875(67.925)+.0062(69.925)= 65.396$
The units above are in amu.
