Electrical – How to i calculate Vs in this circuit knowing Vo=2

The big problem is that you didn't draw the schematic properly. A good schematic always makes the things clear:

^{simulate this circuit – Schematic created using CircuitLab}

The current is \$I = \frac{U}{I}\$

I0 = I1+I2

I0 = (5V-2.065V)/270 = 10.87037037mA

I1 = (2.065V - 1.863V)/100 = 2.02mA

I2 = I0 - I1 = 8.85037037mA

I think that the 1000 in your formula is placed incorrect and probably used when the length is in kilometers. So basically for a one phase load the formula is:

$$V_d = 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L$$ Note: the first 2 is to get the result from source to load and the return as the return cable also influences the voltage drop

For a three phase system the formula is: $$V_d = \sqrt{3} I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L$$

Where:

\$ \begin{align} V_d&=\text{voltage drop in volts}\\ I&=\text{current in amperes}\\ R&=\text{conductive resistance in ohms/m}\\ X&=\text{conductor inductive reactance in ohms/m}\\ L&=\text{one way length of circuit in m (or km/1000 in your formula)}\\ \theta&=\text{phase angle of the load}\\ PF&=\cos(\theta)\\ \end{align}\\ \$

Answer

\$ \begin{align} PF&=1\\ \theta&=\arccos(PF)=0\\ \\ V_d&= 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot\cos(0) + 0.0000704\cdot\sin(0)\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot1 + 0.0000704\cdot0\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001\bigr)\cdot 300\\ V_d&= 2\cdot4.2\\ V_d&= 8.4\\ \\ \end{align}\\ \$

So this gives a voltage drop for one run of 4.2 V and for a double run 8.4 V.

Answer with different PF

Because a PF of 1 in an AC circuit is not a real world example I will show the influence of a PF of 0.8 on the voltage drop:

\$ \begin{align} PF&=0.8\\ \theta&=\arccos(PF)=36.8699°\\ \\ V_d&= 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot\cos(36.8699°) + 0.0000704\cdot\sin(36.8699°)\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot0.8 + 0.0000704\cdot0.6\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.00008+0.00004224\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.00012224\bigr)\cdot 300\\ V_d&= 2\cdot5.13408\\ V_d&= 10.26816\\ \\ \end{align}\\ \$

As a result of the reactive power the voltage drop will increase.