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Ans. A

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Word Access time of $L_1$ = $1ns$ , $L_2 = 5ns$ , $L_3 = 10ns$

As mentioned there is a hit in $L_3$ cache,

Time for transferring block to $L2$ from $L3$ $ = 8 \times 10ns = 80 ns$ ($L_3$ is accessed)

Time for transferring block to $L1$ from $L2$ $ = 8 \times 5ns = 40 ns$ ($L_2$ is accessed)

Time for transferring a word to $processor$ from $L1$ $ = 1ns $ ($L_1$ is accessed)

$121 ns$
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