Assuming a block size of 9,200, 6,000 volumes at the site, and an average of 2 files per volume, the following is the calculation:

1. y1 is 19,800
2. y2 is 6,000 (the number of volumes at the site)
3. y3 is 6,000 * 2 * 2 = 24,000
4. y4 is 10% * 19,800 = 1980
5. y5 is 1980 + 24,000 + 6,000 = 31,980
6. y6 is 31,980 * (49 + 6) = 1,758,900
7. y is (1,758,900 / 9,200) * 2 = 382

num_blks is 1.05 * 382 = 401