Solution

Electric field are vector quantities in this case they are acting in opposite direction.
E  due  to  A$$=\dfrac {k q}{r2}=\dfrac {(9\times 10^9)(50\mu C)}{(0\cdot 3)^2}(\hat x)$$
E  due  to  B$$=\dfrac {(9\times 10^9)(100\mu C)}{(0\cdot 3)^2}(-\hat x)$$
$$\therefore Net \  E=\dfrac {9\times 10^9}{(0\cdot 3)^2}(50-100)\times 10^{-6}\hat x$$
$$E=5\times 10^6(-\hat x)\ V/m$$