Electronic – Define input current of Op-Amp

It's one of the rules. For an ideal opamp goes that

• input impedance is infinite (input current is zero)
• gain is infinite
• output impedance is zero
• input offset voltage is zero

If these requirements aren't met several basic opamp circuits wouldn't work. Take for instance the inverting amplifier.

The transfer function is

\$ \mathrm{V_{OUT} = - \dfrac{R_f}{R_{IN}} \cdot V_{IN}} \$

as derived in this answer. The proof relies on the infinite input impedance, but you can't explain the transfer function based on both inputs equal, because that's not a property of the opamp! So-called proofs that start from the fact that the inverting input is at ground are invalid.

Note that FET input opamps do better than their BJT input counterparts. The former will have pA input current, whereas for the latter this may be several \$\mu\$A.

Further reading:
Opamps for everyone

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 kΩ)/(100 kΩ) = 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 MΩ 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 MΩ) = 10 nA current thru the resistor, which is the opamp input current in this case.