Answer
15.6 mm.
Work Step by Step
Calculate the extreme image distances by using the lens equation with the focal length, and the maximum and minimum object distances.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
$$d_{i,min}=\frac{fd_{o,max}}{ d_{o,max}-f }=\frac{(135mm)(\infty)}{\infty -135mm}= 135mm$$
$$d_{i,max}=\frac{fd_{o,min}}{ d_{o,min}-f }=\frac{(135mm)(1300mm)}{ 1300mm -135mm}= 150.64mm$$
Subtract the extreme image distances, to find the distance $\Delta x$ over which the lens must move (relative to the plane of the sensor or film).
$$\Delta x=150.64mm-135mm=15.6mm$$