The confidence interval is (0.616, 0.686).
To find the confidence interval, we first find p, the proportion of students:
658/1010 = 0.6515
The confidence interval follows the formula

To find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1: 1-0.98 = 0.02
Divide by 2: 0.02/2 = 0.01
Subtract from 1: 1-0.01 = 0.99
Using a z-table (http://www.z-table.com) we see that this is closer to the z-score 2.33.
Using our information, we have:

This gives us the interval (0.6515-0.0349, 0.6515+0.0349) or (0.616, 0.686).