Answer
See the detailed answer below.
Work Step by Step
The electric field direction is toward the charge since it is a negative charge. We chose the origin to be point 1.
And as we see below, the angle $\theta_1$ is in the first quadranet, so
$$\theta_1 =\color{red}{\bf 60^\circ}$$
And since angle $\theta_2$ is in the second quadranet, $$\theta_2=180^\circ -60^\circ =\color{red}{\bf 120^\circ}$$
$\Rightarrow$ At point (1), the electric field magnitude is given by
$$E_1=\dfrac{kq}{r_1^2} $$
Plugging the known;
$$E_1=\dfrac{(8.99\times 10^9)(2\times 10^{-9})}{(0.01)^2} $$
[we neglected the sign of the charge since we are focusing on the electric field direction by angles]
$$E_1=\color{red}{\bf1.8\times 10^5}\;\rm N/C$$
and its direction is $60^\circ$ counterclockwise from $+x$-direction.
$\Rightarrow$ At point (2), the electric field magnitude is given by
$$E_2=\dfrac{kq}{r_2^2} $$
Plugging the known;
$$E_2=\dfrac{(8.99\times 10^9)(2\times 10^{-9})}{(0.01)^2} $$
$$E_2=\color{red}{\bf1.8\times 10^5}\;\rm N/C$$
and its direction is $120^\circ$ counterclockwise from $+x$-direction.
