Atomic mass unit based on other atoms
Question: 


  Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?


Answer

It is easy enough to look up the weights of $\ce{^{12}C}$ (12.00000000...) and $\ce{^9Be}$ (9.0121822) based on $\ce{^{12}C}$ scale and make a decision from that data. Thus a mass scale based on $\ce{^9Be} =9.00...$ would be heavier than an mass scale based on $\ce{^{12}C} = 12.00...$.  

"Justification"

The OP has asserted that the problem is not supposed to be just a "lookup problem" but that some reasoning should be offered. 

I content that the question simply asks which is larger. It doesn't ask why. Furthermore I'd assert that there isn't any "simple" reasoning by which the problem can be analyzed other than lookup. 

Neutron to proton ratio reasoning

In his answer QuarkyLittleThing's reasoned that a neutron is heavier than a proton and since $\ce{^9Be}$ has a higher neutron to proton ratio (5:4) than $\ce{^{12}C}$ (6:6), it ought to be heavier. 

Using the neutron to proton ratio reasoning, $\ce{^9B}$ only has 4 neutrons but 5 protons, so a $\ce{^9B}$ scale should be lighter than the $\ce{^{12}C}$ scale. But the mass of $\ce{^9B}$ is 9.0133288 u, so the neuron to proton ratio isn't entirely sound, and a mass scale based on $\ce{^9B} =9.00...$ would in fact be heavier than a mass scale based on $\ce{^{12}C} = 12.00...$.  

Lower nuclear binding energy reasoning

Mithoron's made a comment on QuarkyLittleThing's answer asserting that the reason was "lower nuclear binding energy." 

To properly analyze Mithoron's comment one must note that the nuclear binding energy is based on the energy differences of the actual fission/fusion of nuclei. Thus as NicolauSakerNeto pointed out on an earlier version of this answer, this is really not a "reason", but a tautology due to $\text{E}=mc^2$. So a decrease in the mass of the nucleus is proportional to an increase in the nuclear binding energy for that nucleus.