Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the charges are uniformly distributed in the disk, the net electric field at a point of $x$ that passes through its center (axis) must be in the $x$-direction.
The $y$-components and $z$-components from the electric field are zeros.
Hence, the net electric field in this situation is then given by
$$E_{disk,x}=\dfrac{\eta}{2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$
where $\eta=\dfrac{Q}{A}=\dfrac{Q}{\pi r^2}$
$$E_{disk,x}=\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$
Note that the electric field from a solid disk is the sum of the electric field from a group of rings and this formula is found by integrating the formula of rind; you can see that in your textbook.
Now we have two positively charged disks, as seen below. So the net electric field at point P just in the middle between them must be zero.
$$E_{net}= (E_1-E_2)\hat i$$
$$E_{net}= \left(\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]-\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]\right)\hat i$$
Noting that $|Q_1|=|Q_2|=Q$, $r_1=r_2=r=5$ cm, and $x_1=x_2=x=10$ cm.
Thus,
$$E_{net}= \color{red}{\bf0}\;{\rm N/C} $$
$$\color{blue}{\bf [b]}$$
Now at the new point K, where $x_1=0.05$ m and $x_2=0.15$ m,
$$E_{net}= \left(\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x_1}{\sqrt{x_1^2+r^2}}\right]-\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x_2}{\sqrt{x_2^2+r^2}}\right]\right)\hat i$$
$$E_{net}=\dfrac{Q}{2\pi r^2\epsilon_0 } \left( 1-\dfrac{x_1}{\sqrt{x_1^2+r^2}} - 1+\dfrac{x_2}{\sqrt{x_2^2+r^2}} \right)\hat i$$
$$E_{net}=\dfrac{Q}{2\pi r^2\epsilon_0 } \left( -\dfrac{x_1}{\sqrt{x_1^2+r^2}} +\dfrac{x_2}{\sqrt{x_2^2+r^2}} \right)\hat i$$
Plugging the known;
$$E_{net}= \dfrac{ (50\times 10^{-9})}{2 \pi (0.05)^2(8.85\times 10^{-12})}\left(-\dfrac{(0.05)}{\sqrt{(0.05)^2+(0.05)^2}}+\dfrac{(0.15)}{\sqrt{(0.15)^2+(0.05)^2}}\right) \hat i$$
$$E_{net}=(\color{red}{\bf 8.7\times 10^4}\;{\rm N/C})\hat i\tag{Rightward}$$