ANS:C - 30 cm

The answer would be 40mm.

Explanation :

Consider the slab of width B with tie bar placed at Centre. The left half of the tie bar will resist the frictional force acting on the left half of the slab (B/2) and the same will happen to the right side.

Taking left half into the picture:

Frictional force = coeff of fric X weight of left half of slab.
= coeff of fric X (B/2)X h X 1m X Unit weight of CC.
= 1.5 x 3.75 x 0.2 x 1 x 2400 = 2700kg.

Resisting force = 1400xArea of steel.
Therefore Area of steel per 1m length = 2700/1400 = 1.92 cm^2.
Spacing = 100 * 3.14 * 1^2/(4x1.92) = 40.88 cm (taking 10mm bar as 1cm in the formula).