Answer
a) $0.03\;\rm N\cdot m^2/C$
b) $0\;\rm N\cdot m^2/C$
Work Step by Step
We know that the electric flux is given by
$$ \Phi=\vec E\cdot \vec A=EA\cos\theta\tag 1$$
And we know that the electric field is uniform over the given area and we know that the area vector is perpendicular to the $xy$-plane.
$$\vec A=0.02\times 0.03$$
$$\vec A=(6\times 10^{-4}\;\hat k)\;{\rm m^2}\tag 2$$
Recall that $\hat k\cdot \hat k=1\times 1\times \cos 0^\circ=1$ and that $\hat i\cdot \hat k=\hat j\cdot \hat k=1\times 1\times \cos 90^\circ=0$
$$\color{blue}{\bf [a]}$$
Plug $\vec E$ from the given formula and $\vec A$ from (2) into (1),
$$ \Phi=(100\;\hat i+50\;\hat k)\cdot (6\times 10^{-4} \;\hat k)=0+(50\times 6\times 10^{-4})$$
$$\Phi=\color{red}{\bf3\times 10^{-2}}\;\rm N\cdot m^2/C$$
$$\color{blue}{\bf [b]}$$
Plug $\vec E$ from the given formula and $\vec A$ from (2) into (1),
$$ \Phi=(100\;\hat i+50\;\hat j)\cdot (6\times 10^{-4} \;\hat k)=0+0$$
$$\Phi=\color{red}{\bf 0}\;\rm N\cdot m^2/C$$
