Single Source Charge
What is the electric field at a distance $d$ from a point charge $q_s$? We can use the tip from the end of Electric Fields and Force: imagine that we place a positive charge $q_t$ at that location, and calculate the force on that star. When we do that, we see that the force on the target is given by [charge/coulombs]: $$F = k{\abs{q_Sq_T}\over r^2}$$ Thus the electric field at the star is $$E = \frac1{q_T} k{\abs{q_S}q_T\over r^2} = k{\abs{q_S}\over r^2}$$ That's the magnitude. For the direction, we see that if $q_S$ is positive, then the positive target charge would move away from it; but if $q_S$ is negative, then the target charge would move towards it. Thus
Multiple Source Charges
If there are multiple source charges, then the total electric field at a point is the sum of the electric field due to each source, added together as vectors.
To find the electric field at this star due to these two sources, we find the electric field due to each source, and then add them together.
The positive charge is $0.4\u{m}$ from the star and has $\abs{q_S}=3\u{\mu C}$, so $$E_+ = k{\abs{q_S}\over r^2} = (9\ten9){(3\mu)\over (0.4)^2} = (9\ten9){(3\ten{-6})\over (0.4)^2} = \mathbf{1.69\ten5\,N/C\;or\;0.169 MN/C}$$ The negative charge is the same distance, but has $\abs{q_S}=2\u{\mu C}$, so $$E_- = (9\ten9){(2\mu)\over (0.4)^2}=1.12\ten5\u{N/C}=0.112\u{MN/C}$$ (Remember M means "mega" or "million": M$=\ten6$.) As for direction, the electric field of the positive charge points away from it, or to the right. (See the second figure; it may be useful to draw in these arrows for yourself.) The electric field of the negative charge points towards it, which in this case is also to the right. Thus the electric fields bolster each other: $$\begin{align} E_{tot} &= 0.169\u{MN/C}\rightarrow + 0.112\u{MN/C}\rightarrow\\ &= \mathbf{0.281\,MN/C\rightarrow} \end{align}$$