The IBM 3350 disk drive uses block addressing. The two subblock organizations aredescribed below.1. Count-data, where the extra space used by count subblock and interblock gapsis equivalent to 185 bytes.2. Count-key-data, where the extra space used by the count, key subblocks andinterblock gaps is equivalent to 267 bytes, plus the key size. An IBM 3350 has 19,069 usable bytes available per track, 30 tracks per cylinder, and555 cylinders per drive. Suppose you have a file with 350,000 byte that you want tostore on a 3350 drive and the record size is 80 byte. Answer the following questions. Unless otherwise directed, assume that the blocking factor is 10 and that the countdata subblock organization is used. a. How many blocks can be stored on one track? How many records?b. How many blocks can be stored on one track if the count-key-data subblockorganization is used and key size is 13 bytes?c. How many cylinders are required to hold the file (blocking factor 10 and countdata format)? How much space will go unused due to internal trackfragmentation?d. If the file were stored on contiguous cylinders and if there were no interferencefrom other processes using the disk drive, the average seek time for a randomaccess of the file would be about 12 msec. Use this rate to compute the averagetime needed to access one record randomly. e. Explain how retrieval time for random accesses of records is affected byincreasing block size Discuss trade-offs between storage efficiency andretrieval when different block sizes are used. Make a table with different blocksizes to illustrate your explanations.