Stalls for instruction always happen when we misses in primary cache
$\therefore for \ 2 \ memory \ reference \ 2 * \frac{1}{5}=0.4 \ misses\ in \ L1 \ cache.$
Similarly for L2 $= 0.2 \ misses$
Now calculate Avg stalls which is $0.4*20+0.2*100=28 \ cycle \ stalls.$