Answer
$I_{20} = 2A$
Work Step by Step
$R_{eq}$ of the circuit is
$R_{eq} =10\Omega$
so current through the circuit is
$ I = \frac{100}{10}A = 10A$
now potential drop across $2\Omega$ and $4\Omega$ is
$V_2 = IR = 10 * 2 V =20 V$
$V_4 = IR = 10 * 4 V =40 V$
since $20\Omega$ and $5\Omega$ are in parallel so potential drop across both will be same and will be equal to
$V_{20} =V_5 = 100V - 20V -40 V =40V$
so current through $20\Omega$ i
$I_{20} = \frac{V_{20}}{R} = \frac{40}{20}A $
$I_{20} = 2A$