Measuring input current of opamp
No, you are not doing this right.  Your concept is good but your calculations are flawed.  The gain of the opamp circuit is not R2/R3, but (R2+R3)/R3.  Your gain is therefore (400 k&Omega;)/(100 k&Omega;) = 4.  I'm using the values in your schematic because I shouldn't have to go looking elsewhere.  If you don't like that, put the real values right on your schematic next time.

You see a change of 4.04V - 3.64V = 400 mV on the output by switching in the 10 M&Omega; resistor.  Divided by the gain of 4, this means a 100 mV change at the opamp positive input.  By Ohm's law, (100 mV)/(10 M&Omega;) = 10 nA current thru the resistor, which is the opamp input current in this case.