Answer
a) $103\;\rm V$
b) $54\;\rm kN/C$
Work Step by Step
First of all, we need to find the points at which the net electric field is zero.
As we see in the figure below, we can see that the net electric field between the two charges is always pointing left and it can never be zero there.
To the right from $q_2$, the electric field from both charges opposes each other; this also happens to the left from $q_1$.
At point $P_1$ where $q_1$ is at the origin where $x=0$ and $q_2$ is at $x=15$ cm.
The net electric field is assumed to be zero at $P_1$,
$$E=-\dfrac{k|q_1|}{r_1^2}+\dfrac{k|q_2|}{r_2^2}\neq 0$$
It is obvious that $|q_2|\gt |q_1|$ and $r_2\lt r_1$, so that the term of $\dfrac{k|q_2|}{r_2^2}$ is always greater than the term of $\dfrac{k|q_1|}{r_1^2}$.
Hence, the electric field will never be zero to the right from $q_2$.
Now let's work with $P_2$,
$$E= \dfrac{k|q_1|}{r_1^2}-\dfrac{k|q_2|}{r_2^2}= 0$$
Hence,
$$ \dfrac{k|q_1|}{r_1^2}=\dfrac{k|q_2|}{r_2^2} $$
$$ \dfrac{|q_1|}{r_1^2}=\dfrac{|q_2|}{r_2^2} $$
where $r_2=0.15+r_1$;
$$ \dfrac{|q_1|}{r_1^2}=\dfrac{|q_2|}{(0.15+r_1)^2} $$
Plug the known;
$$ \dfrac{10}{r_1^2}=\dfrac{20}{(0.15+r_1)^2} $$
Thus,
$$2r_1^2=0.15^2+0.30r_1+r_1^2$$
$$r_1^2-0.30r_1-0.15^2=0$$
So, whether $r_1=-6.2$ cm [which is dismissed since we are dealing with lengths not positions] or $r_1=\bf 36.2$ cm.
This means that the position of $P_2$ is at
$$x_{_{P_2}}=-\bf 36.2\;\rm cm$$
at which the net electric field is zero.
$$\color{blue}{\bf [a]}$$
The net electric potential at $P_2$ is given by
$$V=V_1+V_2=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$$
Plug the known;
$$V= \dfrac{(9\times 10^9)(-10\times 10^{-9})}{(36.2\times 10^{-2})}+\dfrac{(9\times 10^9)(20\times 10^{-9})}{(51.2\times 10^{-2})}$$
$$V=\color{red}{\bf 103}\;\rm V$$
$$\color{blue}{\bf [b]}$$
First, we need to find the point at which $V=0$ V.
$$V=V_1+V_2=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$$
where $r_2=0.15-r_1$;
$$ \dfrac{ \color{red}{\bf\not} kq_1}{r_1}+\dfrac{ \color{red}{\bf\not} kq_2}{0.15-r_1}=0$$
$$ \dfrac{ q_1}{r_1}=-\dfrac{ q_2}{0.15-r_1} $$
Plug the known and solve for $r_1$,
$$ \dfrac{ -10}{r_1}=-\dfrac{ 20}{0.15-r_1} $$
$$2r_1=0.15-r_1$$
$$r_1=5\;\rm cm$$
Hence, the point at which the electric potential between the two point charges is zero is at $x=5$ cm.
Thus,
$$E=-E_1-E_2$$
$$E= -k\left[\dfrac{|q_1|}{r_1^2}+\dfrac{|q_2|}{r_2^2}\right]$$
Plug the known;
$$E= -(9\times 10^9)\left[\dfrac{10\times 10^{-9}}{0.05^2}+\dfrac{20\times 10^{-9}}{0.10^2}\right]$$
$$E=-(\color{red}{\bf 54.0}\;{\rm kN/C})\;\hat i$$
