Answer
$0.14N$ ,rightward
Work Step by Step
There will be a rightward force$F_1$ on $Q_1$ due to $Q_2$ as they are oppositely charged. There will be another force $F_2$towards left on $Q_1$ due to electric field. Let the rightward force be positive.
Thus,
$F_{net}=F_1+F_2$
$\implies F_{net}=\frac{KQ_1Q_2}{x^2}-Q_1E$
We plug in the known values to obtain:
$F_{net}=\frac{8.99\times 10^9(6.7\times 10^{-6}\times 1.8\times 10^{-6})}{(0.47)^2}-6.7\times 10^{-6}\times 53000=0.14N$