Think about what "Efficiency" means and look at the formula.
In general Efficiency is defined as: "what comes out" / "what goes in" x 100%
So a circuit that outputs 0.4 W, while 0.5 W is taken at the input has a 0.4 / 0.5 x 100% = 80 % efficiency.
The formula represents this, the output power: \$I_O * V_O\$
is divided by the Input power: \$(I_O + I_q) * V_I\$ and multiplied by 100%.
Note that this efficiency would always be 100% if \$I_q\$ = 0
This quiescent current \$I_q\$ is the current what the LDO itself consumes.
LDO efficiency calculation output current is we need consider Load current Value or LDO max current rating ?
Would it make sense to use the maximum current rating?
Now think, suppose that \$I_q\$ of the LDO is 0.1 mA.
What would the efficiency be if the load current was also 0.1 mA: \$I_O\$ = 0.1 mA?
What would the efficiency be if the load current was higher, 100 mA: \$I_O\$ = 100 mA?
Hmm, the efficiency is increased when the load current is higher. To me that makes sense as the impact of \$I_q\$ (which is only 0.1 mA) becomes smaller as \$I_O\$ gets larger.
Now do those calculations again but use the maximum current rating \$I_O\$ = 800 mA
Now what happens to these (wrong!) efficiency numbers?
In general: Maximum values are values that you should not exceed, if you do it can damage the device. So your actual values would usually be smaller than the maximum ratings, you have to do your calculations with the actual values, not the maximum ratings.