Its A
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Answer:
 voltage divider,  R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
             R₂ / R₁ = 10³
             R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
 
        
             
        
        
        
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Sorry to say but I know that t(e introduction is first and the coda is last