Since, we know that SI unit of energy is represented by Joule. Hence, let us convert all the different units in the form of Joules.
Option A-
1 Joule= $1\quad Joule\quad =\quad [kg]{ [\frac { m }{ s } ] }^{ 2 }$
Option B-
$kWh\quad =\quad 1000W\times 3600s\quad =\quad 3600000\quad W\times s$
As we know that 1 Ws = 1 Joule
So 1 kWh= 3600000 Joules
Option C-
$1\quad Joule\quad =\quad [kg]{ [\frac { m }{ s } ] }^{ 2 }=\quad \left[ 1000\quad gm \right] \left[ \frac { 100\quad cm }{ 1\quad s } \right] \left[ \frac { 100\quad cm }{ 1\quad s } \right] \ S0,\quad 1\quad Joule\quad =\quad { 10 }^{ 7 }gm\frac { { cm }^{ 2 } }{ { s }^{ 2 } } \quad =\quad { 10 }^{ 7 }\quad erg\ So,\quad 1\quad erg\quad =\quad { 10 }^{ -7 }Joule$
Option D-
Electron Volt is the unit of energy in terms of change in energy for one electron in the potential difference of one volt. So, mathematically it is-
$1\quad ev\quad =\quad Charge\quad of\quad one\quad electron\quad \times \quad 1\quad Volt\quad =\quad 1.6\quad \times { 10 }^{ -19 }Coloumb\quad \times \quad 1\frac { Joule }{ Coloumb } \ So,\quad 1\quad ev=\quad =\quad 1.6\quad \times { 10 }^{ -19 }Joule$
Hence, Option B is the largest unit of energy and hence it is practically used in everyday life of electric consumption.