Solution

For $5$ bus arbiters, we need at least $3$ bits because

$2^2=4<5$

and

$2^3=8\ge 5$

So, the priority encoder must handle $8$ inputs and produce $3$ outputs.

Thus, priority encoder size is

$8 \times 3$

The decoder will take $3$ inputs and produce $8$ outputs.

Thus, decoder size is

$3 \times 8$