Here we have little different version of Arden's Theorem.
If we have $R= PR + Q$ then it has a solution $R = P^*Q$
Proof :
$R= PR+ Q$
$\quad= P(PR+Q)+Q$ (by putting $R= PR+Q$)
$\quad= PPR+PQ+Q$
$\quad=PP(PR+Q)+PQ+Q$ (by putting $R= PR+Q$)
$\quad= PPPR+PPQ+PQ+Q$
and so on, we get $R = \{\ldots+PPPPQ+PPPQ+PPQ+PQ+Q\}$
$\qquad\qquad\qquad= \{\ldots+PPPP+PPP+PP+P+ \lambda \}Q = P^*Q$
or another way:
$R=PR+Q$
$\quad= P(P^*Q) + Q$ (by putting $R = P^*Q$)
$\quad=(PP^* + \lambda )Q = P^*Q$
So, equation is proved.
Now for the Above Question:
$X_1= 0X_1 + 1 \ X_2$ (Equation 2)
$\quad= 0X_1 +1(0X_1 + \lambda)$ ( Put the value of $X_2$ from Equation 3 )
$\quad=0X_1 +10 \ X_1+ 1 = (0+10)X_1 +1$
$X_1= (0+10)^*1$ (Apply if $R = PR + Q$ then $R = P^*Q$)
$X_0 = 1 \ X_1$ (Equation 1)
$X_0 = 1(0+10)^*1$ (Put the value of $X_1$ we calculated).
So, $1(0+10)^*1$.
Option C is correct.