Current Drain on ADC
If you want to express resistor values in 5 significant digits you should at least start with the right voltage:  

\$ \dfrac{115\mathrm{\,V} - 3\mathrm{\,V}}{150 \mathrm{\, \, \mu A}} = 746667 \, \Omega\$,

and the other resistor will be 20 k\$\Omega\$.

Anyway, the values you used will get you 1.5 mA, not 150 \$\mu\$A. And even with 10 times higher resistor values you won't get 3 V, since the input current to your ADC is actually a resistor parallel to the 20 k\$\Omega\$.    

\$ \dfrac{115\mathrm{\,V}  - V_{OUT}}{746.667 \, k \Omega} = \dfrac{V_{OUT}}{20 \, k \Omega} + 150\, \mu A \$

Or \$V_{OUT}\$ = 78 mV. (1.5 mA would leave 2.7 V.)  

Don't feed the divided voltage directly to the ADC, but buffer it first using a voltage follower opamp. Then the 150 \$\mu\$A won't load the divider, and you can safely use higher resistor values, > 1 M\$\Omega\$.

edit (see Telaclavo's comment)
I didn't look at the datasheet at first, I trusted the 150\$\mu\$A was a correct value. ADCs do often have a low input impedance. However, checking the datasheet, Telaclavo is right, the 150\$\mu\$A refers to the reference input (page 3). On page 4 you can see the input leakage current for the analog input: 1\$\mu\$A maximum. That means the ADC has to have a buffer for the input signal.
Anyway, this changes things fundamentally. Suppose you choose the resistors so that they'll draw 100\$\mu\$A: 1.12M\$\Omega\$ + 30k\$\Omega\$.  

\$ \dfrac{115\mathrm{\,V}  - V_{OUT}}{1.12M \Omega} = \dfrac{V_{OUT}}{30 k \Omega} + 1 \mu A \$

Then \$V_{OUT}\$ = 2.97V. That's for a worst case input current of 1\$\mu\$. If you would use an LF411 as buffer, which has an input current of 50pA, the output would be 3V for both the typical leakage current of 1nA, and the maximum of 1\$\mu\$A.