Answer:
[tex](+0.556)\; {\rm nC}[/tex].
Explanation:
The electric field strength at a given position in space is the same as the electrostatic force on a small positive test charge, per unit charge:
[tex]\displaystyle E = \frac{F_{\text{test}}}{q_{\text{test}}}[/tex].
In this question, to ensure that the electrostatic force at point [tex]P[/tex] is zero, the force on a positive test charge at that point should also be zero. In other words, the electrostatic force that [tex]q_{1}[/tex] and [tex]q_{2}[/tex] on this test charge should balance each other.
Electrostatic charges of opposite signs attract each other, while those of the same sign repel each other. Because [tex]q_{1}[/tex] is negative, [tex]q_{1}[/tex] would attract the positive test charge at [tex]P[/tex] towards [tex]q_{1}[/tex] in the negative [tex]x[/tex]-axis direction.
Since electrostatic forces on that positive test charge need to be balanced, [tex]q_{2}[/tex] (which is on the same side of [tex]P[/tex] as [tex]q_{1}[/tex]) would need to repel the positive test charge at [tex]P[/tex]. Hence, the sign of [tex]q_{2}[/tex] should also be positive instead of negative.
Additionally, for the two forces on this test charge [tex]q_\text{test}[/tex] to be balanced, the electrostatic forces that [tex]q_{1}[/tex] and [tex]q_{2}[/tex] exert on this test charge should be equal in magnitude. Let [tex]k[/tex] denote Coulomb's Constant, let [tex]r_{1} = (0.2 - (-0.4)) = 0.6[/tex] denote the distance between [tex]q_{1}[/tex] and [tex]P[/tex], and let [tex]r_{2} = 0.2 - 0 = 0.2[/tex] denote the distance between [tex]q_{2}[/tex] and [tex]P[/tex]:
[tex]\displaystyle \frac{k\, |q_{1}\, q_{\text{test}}|}{{r_{1}}^{2}} = \frac{k\, |q_{2}\, q_{\text{test}}|}{{r_{2}}^{2}}[/tex].
Simplify and solve for [tex]q_{2}[/tex]:
[tex]\displaystyle |q_{2}| = \left(\frac{{r_{2}}^{2}}{{r_{1}}^{2}}\right)\, |q_{1}| = \left(\frac{0.200^{2}}{0.600^{2}}\right)\, (5.00\; {\rm nC}) \approx 0.556\; {\rm nC}[/tex].
In other words, the magnitude of [tex]q_{2}[/tex] should be approximately [tex]0.556\; {\rm nC}[/tex].
Since (based on the analysis above) the sign of the electrostatic charge on [tex]q_{2}[/tex] should positive:
[tex]q_{2} = (+0.556)\; {\rm nC}[/tex].