Correct Answer - Option 1 : 2 pu

__Concept:__

The relation between new per-unit value & old per unit value of reactance
\({({X_{pu}})_{new}}\; = {({X_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({X_{pu}} = \frac{{{X_{Actual}}}}{{{X_{base}}}}\)

\({X_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Xpu)new = New per unit value of reactance

(Xpu)old = Old per unit value of reactance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

__Calculation:__

Given that,

Xd(old) = 1 pu

MVA(new) = 200

MVA(old) = 100

kV(old) = 20

kV(new) = 20

∴ \({X_{d\left( {new} \right)}} = 1 \times \frac{{200}}{{100}} \times {\left( {\frac{{20}}{{20}}} \right)^2} = 2\;pu\)