Line voltage from disc voltages of 4-disc string

The two given disc voltages fix the unknown ratio $m = C_1/C$. Once $m$ is known, the other disc voltages and the line voltage follow.

Relations (discs 1 at tower to 4 at line end)

$$V_2 = V_1(1+m),\qquad V_3 = V_2 + m(V_1+V_2),\qquad V_4 = V_3 + m(V_1+V_2+V_3)$$

Given $V_2 = 13.2$ kV and $V_3 = 18.2$ kV.

Finding $m$

From the first relation, $V_1 = 13.2/(1+m)$. Substituting into the second:

$$18.2 = 13.2 + m\left(\frac{13.2}{1+m} + 13.2\right)$$ $$\begin{aligned} 5(1+m) &= 13.2m(2+m)\\ 13.2m^2 + 21.4m - 5 &= 0\\ m &= \frac{-21.4 + \sqrt{21.4^2 + 4\times13.2\times5}}{2\times13.2} = 0.2072 \end{aligned}$$

Disc voltages

$$\begin{aligned} V_1 &= \frac{13.2}{1.2072} = 10.935\ \text{kV}\\ V_4 &= 18.2 + 0.2072(10.935+13.2+18.2) = 26.971\ \text{kV} \end{aligned}$$

Phase voltage (voltage across the string):

$$V_{ph} = 10.935 + 13.2 + 18.2 + 26.971 = 69.31\ \text{kV}$$

Line voltage

$$V_L = \sqrt3\times 69.31 = 120.04\ \text{kV}$$

Answer: The disc voltages are 10.94, 13.2, 18.2 and 26.97 kV, the string voltage is 69.31 kV, and the line voltage is about 120 kV.