The $1$st octet is all $1$s$,$ the $2$nd octet is all $1$s$,$ and the $3$rd octet $41$s and $40$s$.$ Therefore, the $3$rd octet is

Another way to determine this is by looking at the first decimal number which is $172$ class B address therefore the next octet is the $3$rd one to be used for sub$-$ netting.

$2.$ The decimal number of the $3$rd octet in the subnet mask is $240($or $11110000).$ Therefore the address block size is $255-240=16.$ That means there will be $16$ subnets in total $($remember we don$$t exclude the $0000$ and $1111$ combinations of the borrowed for sub$-$netting bits$-$ The number of subnets $=24=16)$

$3.$ The first subnet is $172\mathrm{.}16\mathrm{.}0\mathrm{.}0(172\mathrm{.}16\mathrm{.}0\mathrm{.}0/20$ with the $4$ borrowed bits in the $3$rd octet set to $0)$

$4.$ Beginning with the first subnet of $172\mathrm{.}16\mathrm{.}0\mathrm{.}0/20$ and counting the block size of $16$ in the interesting octet yields

the following subnets: