Ok. I think I found out how to do this.

First calculate $\displaystyle P^3$

Then

$\displaystyle Pr(X_3=0)=(\frac{1}{4})(\frac{39}{108})+(\frac{1}{ 4})(\frac{48}{108})+(\frac{1}{2})(\frac{45}{108})= \frac{177}{432}$

$\displaystyle Pr(X_3=1)=(\frac{1}{4})(\frac{22}{108})+(\frac{1}{ 4})(\frac{16}{108})+(\frac{1}{2})(\frac{24}{108})= \frac{86}{432}$

$\displaystyle Pr(X_3=2)=(\frac{1}{4})(\frac{47}{108})+(\frac{1}{ 4})(\frac{44}{108})+(\frac{1}{2})(\frac{39}{108})= \frac{169}{432}$

Then $\displaystyle E(X_3)=(0)(\frac{177}{432})+(1)(\frac{86}{432})+(2 )(\frac{169}{432})=\frac{424}{432}$

Can someone confirm whether this is correct or not? Thanks