I don't think a consideration of wet/dry leaves or anything like that is necessary.

Efficiency, \(\displaystyle e\), is defined as:

\(\displaystyle e = \frac{total \, used \, energy}{total \, input \, energy} = \frac{E_{use}}{E_{input}}\)

So we need to evaluate the two energies for the ten year period and then divide one by the other to get the efficiency.

The total input energy is going to be the total solar radiation captured by the tree. You know the incident solar irradiance (168 W/m2) so if you multiply this value by the capture area (the area of a circle of radius 8m) then you'll get the total input power, P:

\(\displaystyle P = A I_{solar} = \pi r^2 I_{solar}\)

Now, power is the rate of change of energy, so to get the total energy gathered over a year, you need to multiply the power by the time (in seconds).

\(\displaystyle E_{input} = P t\)

where \(\displaystyle t\) is the duration of 10 years (in seconds).

You might also want to consider the dormancy period here... since the time spent growing is actually less than ten years, you can calculate the duration spent actually growing and use that for \(\displaystyle t\) instead.

What about day-time and night-time? Maybe we can ignore this for now and then investigate it later

To get the useful energy, \(\displaystyle E_{use}\), we need to consider how much energy is gathered and actually put into chemical energy. There's different ways of calculating this, but whatever you do, don't use \(\displaystyle E=mc^2\)... you'll get a massive number that's very unrealistic.

I found a book that said that glucose stores 16000 J of chemical energy per gram. This is similar to wood, which is about 15000 J per gram (from wikipedia). It's up to you which value you want to use. If we take 15000 J of chemical energy per gram, then that's 15 MJ per kg of energy. It should be straightforward to then get the energy wrapped up in 540 kg of wood.

Once you have the two energies, you can get the efficiency