No. Power generally is proportional to the voltage squared (P∝U²). P=U²/R → U=√(PR)

The numbers you quote (1W [not dBm!] is 7.1V) make me believe that you assume an impedance of 50Ω.

    In [1]: math.sqrt(1*50)     # P[in W] * R[in Ohm]
    Out[1]: 7.0710678118654755  # U[in V]

So, in an hypothetical cable of 50Ω impedance, 50kW of RF power is...

    In [2]: math.sqrt(50e3*50) # P[in W] * R[in Ohm]
    Out[2]: 1581.1388300841897 # U[in V]

about 1.6kV(RMS).

Note, though, that with large installations it is likely that there are a few individually tuned feeds. And the cable might not be a 50 Ohm cable, but maybe a chicken ladder (two strands separated by fixed spacers), large-area waveguide, ...

eh i wasnt expecting to be right. Although i would say they usually use heavy duty coax at 50ohm - i believe - not sure though

The pictured parabola antenna likely has a gain around +20dB. Would that change your calculation?

i don't think it would change (probably wrong) since the active energy is only generated by the final PA's in the system. the Antenna's gain is only from the passively reactive design. Its like making a violin string resonate by only moving the bow and not physically the string.

i mis-read your comment - it has no affect at all as its not whats transmitting 50kW!