An L1 cache has four-word blocks and 2 ns access time; L2 has sixteen-word blocks and 20 ns access time; main memory takes 200 ns. Both buses transfer four words at a time. If both caches miss, how long does it take to fill the L2 block and then the required L1 block?
The sixteen-word L2 block needs four transfers of four words. Each main-memory read plus L2 write takes 200+20=220 ns, totalling 880 ns. Then one four-word transfer from L2 into L1 takes 20+2=22 ns. Total is 902 ns, C. The measurement is the block-fill process stated, not an additional initial lookup.
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