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Solution

$T_t=333.33\mu sec$

$T_p=18000\mu sec$

Now,

$a=\frac{T_p}{T_t}$

$a=\frac{18000}{333.33}$

$a=54$

For maximum efficiency in sliding window protocol, sender window size should be:

$W=1+2a$

$W=1+2(54)$

$W=109$

So, sender window size required for maximum efficiency is $109$.