IP address of the computer =$200.99.39.112$
Subnet mask=$255.255.225.224$
Subnet mask is 32 bit dotted decimal represented number created by setting all network bit to $1$ and all host id bit to $0$.
so here subnet mask =$11111111.11111111.11111111.11100000$ which implies we have $27$ bit as network bit and $5$ bit as host id.
Given IP address belongs to class C network which has $24$ bit as network id and $8$ bit as host id .But subnet mask we have $27$ bit bit as network id and $5$ bit for host id . So it implies we use $(27-24)=3$ bit for subnetting which gives us $2^{3}=8$ subnet.
Now we need network IP address for the given IP address for that we need to do bit wise and with IP address and Subnet mask.
$200.99.39.112$
$255.255.255.224$
$200.99.39.96$
So if we expand the last octet in binary we see
$01100000$ ---- It represent 4th subnet network id as subnet bit is $(011)_{2}$=$3_{10}$ , It is 4th subnet as first subnet is $(000)_{2}$=$0$ ,second subnet is $(001)_{2}$=1 so on.
Now we need 6th subnet for that subnet bits would be 101.
So the network id of the 6th subnet is
$200.99.39.160$ -----[$160_{10}$=$10100000_{2}$]
Last assignable ip address for this subnet is when all the host id is 1 except the last bit so that is should not be broadcast address for this subnet .
so The last assignable IP address for the 6th subnet is
$200.99.39.190$.
So the decimal value of last octet of last host of sixth subnet is $190$.