I think regular expression 1 is correct.
$\because \color{blue} {a^*(ba^*)^*=b^*(ab^*)^*=(a+b)^*}$
And one more important conversion--
$\color{cyan}{(PQ)^*P=P(QP)^*}$
From this we can write $a^*(ba^*)^*=(a^*b)^*a^*$
where P is a* and Q is b.
So in given question--
$0^*(10^*)^*=(1+0)^*=1^*(01^*)^*=(1^*0)^*1^*$
Option 1 is correct..