Answer
since the $y$ -component of each force will cancel, the net force points in the $-x$ direction, with a magnitude
$$
2 F_{x}=2\left(G m M r^{2}\right) \cos \theta
$$
where $$\theta=\tan ^{-1}(4 / 3)=53^{\circ} $$ Thus, the result is $$\vec{F}_{\text {net }}=\left(-6.4 \times 10^{-8} \mathrm{N}\right) \hat{\mathrm{i}} $$
Work Step by Step
since the $y$ -component of each force will cancel, the net force points in the $-x$ direction, with a magnitude
$$
2 F_{x}=2\left(G m M r^{2}\right) \cos \theta
$$
where $$\theta=\tan ^{-1}(4 / 3)=53^{\circ} $$ Thus, the result is $$\vec{F}_{\text {net }}=\left(-6.4 \times 10^{-8} \mathrm{N}\right) \hat{\mathrm{i}} $$
