@Arjun Sir
We shouldnot consider only numerator part
They are asking about $Pr[Y=0 \mid X_3 = 0] $ ,right?(Not only numerator part)...........(i)
Now, we recall the formula $Pr[A \mid B] =\frac{Pr\left ( A\cap B \right )}{Pr\left ( B \right )}$
Put this formula in (i)
We get,$\frac{Pr[Y=0 \cap X_3 = 0]}{Pr[X_{3}=0]}$.......................(ii)
Now putting the value of $Y=X_{1}X_{2}\oplus X_{3}=\left \{ X_{1}X_{2}X_{3}' +\left ( X_{1}X_{2} \right )'X_{3}\right \}$ in the equation ii we get
$$\frac{Pr[\left ( \left \{ X_{1}X_{2}X_{3}' +\left ( X_{1}X_{2} \right )'X_{3}\right \}=0 \right )\cap \left ( X_3 = 0 \right )]}{Pr\left ( X_{3} =0\right )} $$ .........................iii
Now, as As $X_3 = 0$ is given, to have Y = 0, $X_1X_2$ should be $0$, meaning $\left(X_1, X_2 \right)$ should be one of $\left\{(0,0) (0,1) (1,0)\right\}$
So, There are totally 4 possibilities - $\left\{(0,0) (0,1) (1,0), (1,1)\right\}$, out of which 3 are favourable cases.
So, X_{1}X_{2} =3/4 and ( X_{1}X_{2})'=1/4
and X_{3}=1/2 and X_{3}'=1/2
Now, putting it in above equation (iii)
$\frac{ 3/4 * 1/2 + 1/4 * 1/2}{1/2}=1/2$