The two possible solutions are as follows.
The first diagram is Ra in parallel with R2. Then R3 is in series with the result. Finally, R1 is in parallel with the result.
Ra || R2 = 0.99
0.99 + R3 = 10.99
10.99 || R1 = 5.15
Notice that in the second diagram the voltage of the supply is lower.
This is called a Thevenin equivalent.
For a Thevenin equivalent first, remove the load and calculate the voltage across where the load was. In this case, R2 and R3 form a resistive divider.
V1 * (R2 / (R3 + R2)) = 0.090V
Then second, short the supply (turn it off or reduce it to zero for a voltage supply and calculate the equivalent resistance. With the supply sorted R3 is completely bypassed. This leaves R2 in parallel with R1.
R1 || R2 = 0.9 ohms.
Finally, take the voltage and resistance and set it up in the manner shown.
simulate this circuit – Schematic created using CircuitLab
To keep current the same though Ra first you need to calculate what the current in Ra is. Then find the resistance needed to keep the same current. First Ra and R2 are in parallel.
R2 || Ra = .99
Then the result is in series with R3. (R1 is ignored because it is directly across the supply and does not affect Ra)
R3 + 0.99 = 10.99
The current is then V1 / 10.99 = 0.090A
Then you do a current divider to see what part of the current is flowing through Ra and what part is flowing through R2.
0.090 * (R2 / (Ra + R2)) = 9.09e-5
Now set up an equasion with X equal to the needed resistance. The total voltage drop in the circuit should be 1V.
9.09e-5 * 1000 + 9.09e-5 * X = 1
Solve for X to get 10121 ohms.
simulate this circuit