How significant is the broadening of a nsec pulse laser in the optical fibers? Their fiber was quartz, 2

Aperture a = 0.22 in air means that the angle A between the axis of the fiber and the ray most inclined from the axis IN AIR has sin A = a

The precise value of the index of refraction of quartz at wavelength 337 nm, should be found in your textbook. I shall take it n = 1.5 for the example here, but if you find a more precise value in your books, use the more precise value.

From the Snell's law, the angle B between the axis of the fiber and the ray most inclined from the axis IN FIBER has

sin B = sin A / n = a/n

A ray following the axis of the fiber exactly travels with speed

v0 = c/n

where the speed of light in vacuum is c = 3e8 m/s.
On the other hand, a ray traveling at angle B to the axis (bouncing from the walls with internal reflection) travels with speed

v1 = v0*cos B

The times needed to travel the length L = 2.5 m of the fiber for the two rays are

t0 = L/v0 = nL/c

and

t1 = L/v1 = nL/(c cos B) = (nL/c) * n/sqrt(n^2-a^2),

respectively.
These are the minimum and maximum times requested in the assignment.

The temporal pulse broadening is the difference between these two times:

Delta t = t1-t0 = (nL/c) * [ n/sqrt(n^2-a^2) - 1 ]

= 1.5 * 2.5 m / (3.e8 m/s) *( 1.5/sqrt(2.25 - 0.0484) - 1) = 1.4e-10 s = 0.14 ns

If the original pulse lasted 1 ns, it will be broadened by 14%.