Lets consider for Tavg-1.
Avg access time for L2
= L2 hit ratio * (L2 access time)
+ L2 miss ratio * ( L2 search to conclude miss + Avg L3 access time + L2 access to accomodate new data)
= 0.9 * (100) + 0.1 * (100 + 4*500 + 4*100)
.....(we multiply by 4 because for each word transfer, 1 mem cycle is consumed)
= 340 ns
Avg access time for L1
= L1 hit ratio * (L1 access time)
+ L1 miss ratio * ( L1 search to conclude miss + Avg. L2 access time + L1 access to accomodate new data)
= 0.8 * (2) + 0.2 * (2 + 2*34 + 2*2 )
...............(we multiply by 2 for the same reason above)
= 16.4ns
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Now, we consider Tavg-2.
Here, the data width=4 and hence, 1 mem cycle is suficient for transfer of data between any two levels of memory.
Thus, Avg L2 access time
= 0.9 * (100) + 0.1 * (100 + 1*500 + 1*100)
= 160ns
Avg access time for L1
= 0.8 * (2) + 0.2 * (2 + 1*34 + 1*2 )
= 9.2ns
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Tavg-1 / Tavg-2
= $\frac{16.4}{9.2}$
= 1.78