Reactance diagram of eight-bus system

Choice of base: 10 MVA, with 11 kV in the generator circuits (the transformer LV rating) and 66 kV on the lines. The motor rating is not printed, so its reactance is left as $X_M$ (pu on its own rating). Its value on the common base is $X_M\frac{10}{S_M}\left(\frac{V_M}{11}\right)^2$.

$$Z_{b,66} = \frac{66^2}{10} = 435.6\ \Omega$$

Per-unit reactances (10 MVA base)

$$\begin{aligned} X_{G1} = X_{G2} &= 0.10\times\frac{10}{10}\times\left(\frac{11.5}{11}\right)^2 = 0.1093\ \text{pu}\\ X_{G3} &= 0.12\times\frac{10}{15}\times\left(\frac{11.5}{11}\right)^2 = 0.0874\ \text{pu}\\ X_{T1} = X_{T2} &= 0.10\ \text{pu}\\ X_{T3} &= 0.15\times\frac{10}{15} = 0.10\ \text{pu}\\ X_{T4} &= 0.18\times\frac{10}{35} = 0.0514\ \text{pu}\\ X_{L1} = X_{L2} = X_{L3} = X_{L4} &= \frac{20}{435.6} = 0.0459\ \text{pu} \end{aligned}$$

Reactance diagram

 (E1)-j0.1093-[1]-j0.10-[2]---j0.0459(L2)---[6]
                         |                   |
                   j0.0459(L1)          j0.0459(L4)
                         |                   |
 (E2)-j0.1093-[3]-j0.10-[4]---j0.0459(L3)---[7]

 [6]-j0.10(T3)-[5]-j0.0874-(E3)
 [7]-j0.0514(T4)-[8]-jXM-(Em)

 All sources (E1, E2, E3, Em) return to the
 common reference (neutral) bus.

Answer: On 10 MVA (11 kV / 66 kV bases): $X_{G1} = X_{G2} = 0.1093$, $X_{G3} = 0.0874$, $X_{T1} = X_{T2} = X_{T3} = 0.10$, $X_{T4} = 0.0514$, and each line 0.0459 pu.