Answer is $0.75$
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, required probability $= 3 \times \dfrac{1}{2} \times \dfrac{1}{2} = 0.75 \because $ we can choose any of the $3$ possibilities in $3$ ways and then probability of each set of two combination is $ \dfrac{1}{2} \times \dfrac{1}{2}$.
We can also do like follows:
There are totally $4$ possibilities - $\left\{(0,0) (0,1) (1,0), (1,1)\right\}$, out of which $3$ are favourable cases.
So, required probability $ = \dfrac{3}{4} = 0.75$.