Answer
There is one answer, at x = 4.3 m.
Work Step by Step
The electric field strength drops off quickly. Any place where the field is zero is closer to $Q_2$, since it is weaker. In between the two charges, the fields due to the two charges both point to the right, and cannot cancel. Thus the only possible answers lie at points greater than $x_2$ = 2.4 meters. Let the proposed location be at x = L.
The fields will cancel, so set the magnitudes equal.
$$k\frac{Q_1}{L^2}= k\frac{Q_2}{(L-x_2)^2}$$
$$\frac{2.5\times10^{-5}}{L^2}= \frac{5.0\times10^{-6}}{(L-2.4)^2}$$
Solving, L = 4.3 meters.
