Answer
See the detailed answer below.
Work Step by Step
First, we need to sketch the given graph to draw the electric fields directions, as shown below.
$$\color{blue}{\bf [a]}$$
$\bullet\bullet$ The electric field at point P from charge $q_1$ is given by
$$E_1=\dfrac{kq_1}{r_1^2}(\cos\theta_1\;\hat i+\sin\theta_1\;\hat j)$$
where, from the geometry of the figure below, $\sin\theta_1=-0.01/r_1$, and $\cos\theta_1=0.03/r_1$.
So,
$$E_1=\dfrac{kq_1}{r_1^3}(0.03\;\hat i-0.01\;\hat j)$$
and we can see that $r_1=\sqrt{0.01^2+0.03^2}$
$$E_1=\dfrac{kq_1}{(0.01^2+0.03^2)^{\frac{3}{2}}}(0.03\;\hat i-0.01\;\hat j)$$
Plugging the known;
$$E_1=\dfrac{(9.0\times 10^9)(1\times 10^{-9})}{(0.01^2+0.03^2)^{\frac{3}{2}}}(0.03\;\hat i-0.01\;\hat j)$$
$$E_1=(\color{red}{\bf 8.5}\;\hat i-\color{red}{\bf 2.8}\;\hat j)\;\rm kN/C$$
$\bullet\bullet$ The electric field at point P from charge $q_2$ is given by
$$E_2=\dfrac{kq_2}{r_2^2}(\cos\theta_2\;\hat i+\sin\theta_2\;\hat j)$$
where, from the geometry of the figure below, $ \theta_2=0^\circ$
So that,
$$E_2=\dfrac{kq_2}{r_2^2}(1\;\hat i+0\;\hat j)=\dfrac{kq_2}{r_2^2} \;\hat i$$
Plugging the known;
$$E_2=\dfrac{(9.0\times 10^9)(1\times 10^{-9})}{0.03^2} \;\hat i$$
$$E_2=(\color{red}{\bf 10}\;\hat i )\;\rm kN/C$$
$\bullet\bullet$ The electric field at point P from charge $q_3$ is given by
$$E_3=\dfrac{kq_3}{r_3^2}(\cos\theta_3\;\hat i+\sin\theta_3\;\hat j)$$
where, from the geometry of the figure below, $\sin\theta_3=0.01/r_3$, and $\cos\theta_3=0.03/r_3$.
So,
$$E_3=\dfrac{kq_3}{r_3^3}(0.03\;\hat i+0.01\;\hat j)$$
and we can see that $r_3=r_1=\sqrt{0.01^2+0.03^2}$
$$E_3=\dfrac{kq_3}{(0.01^2+0.03^2)^{\frac{3}{2}}}(0.03\;\hat i+0.01\;\hat j)$$
Plugging the known;
$$E_3=\dfrac{(9.0\times 10^9)(1\times 10^{-9})}{(0.01^2+0.03^2)^{\frac{3}{2}}}(0.03\;\hat i+0.01\;\hat j)$$
$$E_3=(\color{red}{\bf 8.5}\;\hat i+\color{red}{\bf 2.8}\;\hat j)\;\rm kN/C$$
$$\color{blue}{\bf [b]}$$
First, we need to recall that the electric field is a vector quantity like the force.
It is given by $E=qF$, so it had the same principels of adding forces.
Hence, it must obeys the superposition principle.
So the answer is $\bf yes$, it obeys the principle of superposition.
$$\color{blue}{\bf [c]}$$
We just have to add these three cector quantities and since they are in a component form, it is too easy.
$$\vec E_{net}=\vec E_1+\vec E_2+\vec E_3$$
$$\vec E_{net}= 8.5 \;\hat i- 2.8 \;\hat j + 10\;\hat i+8.5 \;\hat i+ 2.8 \;\hat j$$
The net $y$-component is zero.
$$\vec E_{net}=(\color{red}{\bf 27}\;\hat i)\;\rm kN/C $$
