In this question nothing is mention about search time inside memory and disk, so we safely ignore search time .
so that average access time is :
cache access time + MM access time + Disk access time
$$\begin{align*} & 0.8 * 30 \text{ ns } \\ &+ 0.18 * \left ( 30 + 80 \right ) \text{ ns } \\ &+ 0.02 * \left ( 30 + 80 + 22000000 \right ) \text{ ns } \\ &=\bf 440046 \text{ ns } \\ \end{align*}$$
in each above case , cache and MM and disk we consider search time to find that referenced word = 0 .