Let $C$ is the $TLB$ access time and $M$ is main memory access time.
So Effective memory access time ($EMAT$) in multilevel cache,
$EMAT\ =h(C+M)+(1-h)(C+(n+1)M)$
Here no of level $n$=4 So,
$EMAT\ =h(C+M)+(1-h)(C+5M)$
Given here $h=.98$ , Main memory access time ($M$)=100 ns and
$TLB$ access time ($C$)=20ns.
$EMAT\ =.98(20+100)+.02(20+5\times 100)$
$=.98(120)+.02(520)$
$=128ns$
Reference:Multilevel paging