Explanation :
Base current,
$ \begin{array}{l}\;\;\;{\mathrm I}_\mathrm B=\frac{30\times10^3}{\sqrt3\times13.8}=1255.109\;\mathrm A\\\;\;\;{\mathrm l}_\mathrm f=4270\;\mathrm A\\{\mathrm l}_{\mathrm p.\mathrm u.}=\frac{4270}{1255.109}=3.402\;\mathrm p.\mathrm u\\\;\;{\mathrm l}_{\mathrm g1}=\frac{{\mathrm E}_\mathrm a}{{\mathrm X}_1+({\mathrm X}_2\vert\vert{\mathrm X}_0)}\end{array} $
Where $ {\mathrm X}_0={\mathrm X}_0+3({\mathrm Z}_\mathrm n+{\mathrm Z}_\mathrm f) $
$ \begin{array}{l}\;\;\;\;{\mathrm X}_0=3{\mathrm Z}_\mathrm n+0.35\\3.402=\frac{1.0}{0.15+{\displaystyle\frac{0.15\times(3{\mathrm Z}_\mathrm n+0.35)}{0.15+3{\mathrm Z}_\mathrm n+0.35}}}\end{array} $
by solving the equation
$ \;\;\;\;{\mathrm Z}_\mathrm n=1.07\;\mathrm p.\mathrm u. $