Let the addresses of two consecutive bytes in main memory be $(E201F)_{16}$ and $(E2020)_{16}$
$E20$ is common to both hex numbers, so ignore it.
$1F=0001 \ 1111=31 \ (decimal)$
$20=0010 \ 0000=32 \ (decimal)$
So, difference between two consecutive memory locations = $1 \ B$ Hence, the memory is byte addressable.
First $4$ bits would be tag bits, and next $12$ bits would be index bits.
So, for $(E201F)_{16}$
Tag = $E$; Line = $201$