Electrical power is given by P=RV2⇒R=PV2

Resistance of lamp is R=50100×100⇒R=200Ω
Let VC and VR be the voltage across capacitor and resistor.
Here, VR2+VC2=V2
Current through lamp is i=200100=21A
⇒(100)2+VC2=(200)2
⇒VC2=30000⇒VC=1003V
We know that, V=i×XC
Then, XC=2003Ω=ωC1
So, capacitance of capacitor is C=20×50×2031=x50×10−6
x=50×10−6×100×2003
Thus, value of x=3.