A step-index optical fibre has a core of refractive index 1.62 and a cladding of refractive index…

Step 1. The numerical aperture. $\sin\theta_a = \sqrt{n_1^2 - n_2^2} = \sqrt{2.6244 - 2.3104} = \sqrt{0.314} = 0.560$.

Step 2. The acceptance angle. $\theta_a = \sin^{-1}(0.560) = 34.1°$.

Step 3. The critical angle in the fibre. $\sin C = \dfrac{1.52}{1.62} = 0.938$, so $C = 69.8°$.

Step 4. The two paths. The ray along the axis travels $L = 1000\ \text{m}$. The steepest trapped ray makes the angle $90° - C$ with the axis, so its path is longer by the factor $\dfrac{1}{\sin C} = \dfrac{n_1}{n_2} = 1.066$. Its path is $1066\ \text{m}$.

Step 5. The time difference. The speed in the core is $\dfrac{c}{n_1} = \dfrac{3.00 \times 10^{8}\ \text{m/s}}{1.62}$. The difference of the paths is $\Delta s = 1000\ \text{m} \times \left(\dfrac{1.62}{1.52} - 1\right) = 65.8\ \text{m}$. So $\Delta t = \dfrac{\Delta s \times n_1}{c} = \dfrac{65.8\ \text{m} \times 1.62}{3.00 \times 10^{8}\ \text{m/s}} = 3.55 \times 10^{-7}\ \text{s} = 355\ \text{ns}$.

What it means. A pulse that is short when it goes in is about $355\ \text{ns}$ longer after $1.00\ \text{km}$. Pulses that follow each other more closely than this run together. A thinner core, or a core whose index changes smoothly, cuts this spreading.