"Manipulating the circuit" involves choosing resistance values for \$R_1\$ and for \$R_2\$

ah so im tasked to play with the resistance? using this relation \$V_o = -\frac{R_f}{R_l} v_1 \$ but do i substitute with the expression containing the AC signal? and how does the maximal current appear in such expression? @glen_geek

The signal current available gives you a lower limit on the value of R1. The inverting gain is indeed R2/R1. The upper limit is given by noise in the resistors, and the bias current required at the opamp input. Get some plausible values from somewhere, or just guesses for those, and state those as assumptions when giving your answer.

how would i extract the signal current expression though ? @Neil_UK

Do you see the virtual ground?

yes at (-) but why? @SpehroPefhany

Does that help you calculate the input current?

but we have an alternating voltage input? or do we say that \$ R_1 = \frac{V_1}{I_{1_max}} = \frac{0.1}{5\cdot 10^{-6}} , 5R_1 = R_2 \$ @SpehroPefhany

Think of the peak current from the sinusoidal input voltage.

For finding two unknowns (R1 & R2), you need two equations to solve. One equation involves voltage gain, the other involves \$5 uA\$ flowing from \$v_I\$.

@glen_geek can you look at the edit i made into the post regarding the solution?

@MOHAMEDSALHI Looks like the virtual ground that Spehro mentioned is still eluding you. Otherwise, I would think your solution is right at your fingertips. Do you feel you understand his point when he wrote the word virtual? It's important.

hello @periblepsis , no i dont think i understand the point he is trying to prove, is what i wrote in the post wrong?

@SpehroPefhany the peak current from the sinusoidal input voltage is \$I = \frac{0.1}{R_1} = 5 \mu A \$ no?

That is correct, so you know the minimum value for R1.

yes and i use that to calculate \$ R_2 \$ ? @SpehroPefhany and that will conclude the problem?

You've pretty much arrived at a solution, but all of us ask you to examine how this circuit manages AC current flow and AC voltage so that opamp "-" input pin is a virtual ground. This is an important aspect to comprehend and will aid your understanding of many other opamp circuits.

@glen_geek yes thats what i fail to grasp any answer to that question will be an extreme blessing to my knowledge , infinite gratitude!

Yes, that would yield one correct answer. Of course R1 could be higher than the minimum with some kind of upper limit as Neil_UK has mentioned. Also, as periblepsis & glen_geek say, make sure you understand why the virtual ground is at 0V. It's important and will trip you up later if you don't.

@SpehroPefhany i dont really understand why we imposed that to be 0V , care to explain it please?

The op-amp is very high gain, which drives the inputs very close to equal with negative feedback. Check your class notes, this should have been covered.

ah you want to tell me why V- = 0V? because its similar in someway to a MOSFET there is a infinite resistance between (-) and (+) and somewhat we have gate-like behavior so the current inside it is zero @SpehroPefhany