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