Electrostatics: JEE Main 2019 Physics question

Two point charges q₁$\left( {\sqrt {10} \mu C} \right)$ and q₂($-$ 25 $\mu$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, [take ${1 \over {4\pi { \in _0}}}$ = 9 $\times$ 10⁹ Nm²C^$-$2]