Two small beads having charges q1 and q2 of the same sign are fixed at the opposite ends of a horizontal insulating rod of length
d. the bead with charge q1 is at the origin. as shown in the figure below, a third small, charged bead is free to slide on the rod.

There is no diagram, so I'll assume q1 is at x=0 and q2 is at x=1.5 m. 

Call the 3rd (positive) charge Q and take its equilibrium position as x=X. 

Distance from q1 to Q is X 
Force on Q due to q1 = k(17q)Q/X² acting to the right (see link) 

Distance from q2 to Q is 1.5-X 
Force on Q due to q2 = k(q)Q/(1.5-X)² acting to the left (see link) 

When Q is in equilibrium these 2 opposite direction forces must have equal magnitudes: 

k(17q)Q/X² = k(q)Q/(1.5-X)² 
17/X² = 1/(1.5-X)² 

Square root both sides 
4.123/X = 1/(1.5-X) 

4.123(1.5-X) = X 
6.185 - 4.123X = X 
5.123X = 6.185 
X = 1.21m