Let the inner and outer radii of the track be r meters and R meters respectively.

Then,
= 2πr = 528 m
= 2 × (22/7) × r = 528
= r = (528/ (2 × (22/7)))
= r = (528/ 2) × (7/22)
= r = (264/2) × (7/11)
= r = 132 × 0.6364
= r = 84 m
And,
= 2πR = 616 m
= 2 × (22/7) × R = 616
= R = (616/ (2 × (22/7)))
= R = (616/ 2) × (7/22)
= R = (308/2) × (7/11)
= R = 154 × 0.6364
= R = 98 m
Now,
(R – r) = (98 – 84)
= 14 m
Hence, the width of the track is 14 m