Explain to me this solved part of exercise please?
I think it must be a typo. Using a completely different approach, I came up with a value for Rin of 766K.

Assuming &micro; is really 104 (not 104), I solved for the voltage at the inverting input of the opamp as a function of V1, and came up with V- = V1 &times; 6.6555. This means that the effective resistance between the inverting input and ground (the real value of R11) must be 6.6555 &times; Rid, or 665.55 k&Omega;. Rin is the sum of these two resistances, or 765.55 k&Omega;.