Answer
$28.0854$ amu
Work Step by Step
We simply take a weighted average:
$w_{28}m_{28}+w_{29}m_{29}+w_{30}m_{30}=M$
where $w_{n}$ is the atomic average and $m_{n}$ is the atomic weight of the specific silicon isotope with mass number $n$. M is the final average weight that we wish to verify.
$.9223(27.9769)+.0468*(28.9765)+.0309(29.9738)=$$ 28.0854$ amu
Hence, the atomic weight has been confirmed.
