solution: part b.

Again we start with a drawing of the fields felt by a positive test charge because of the two charges:

Once again we see that the distances (r) from the test charge to the charges are equal, so the magnitudes will be equal:

Now the directions of the two fields point away from each charge starting at the test charge, so we need to add them up. As you can see from the picture, the x components of the fields cancel each other, so you just need to add the y components together to find the final electric field.

This is directed straight up in the y-direction.