Credit goes to original poster found it so thought shuld let u know too.
As the two points are on the X axis and the electric field is along positive X axis therefore the position must be on the X axis.
let rA = x-5 from point A= (5,0) where x is original position of the charge
and rB = x-10 from point B=(10,0)
we know EA = 10N/C and EB = 15N/C
Then EB/EA = 1.5
But E directly proportional to (1/r2)
∴(rA/rB)2 = 1.5
(x-5)2 /(x-10)2 =1.5
x2+25-10x = 1.5(x2+100-20x)
.5x2 -20x + 1252= 0
Then x = 32.247cm or 7.753cm
But the direction of electric field is same at both points. therefore x = 7.753cm will not give the desired result. therefore x = 32.247cm
Therefore the point is (32.247cm, 0)
(b)
As the charge is beyond the points and the direction is along positive X axis, therefore the direction of the electric field is towards the point charge. therefore the sign of the charge is negative.
Now EB = 15N/C
( 8.99*109Nm2/C2)(q)/(32.247cm-10cm)2 = 15N/C
q = .8249*10-12C
∴charge is q = -0.8249μC