$\text{Directed Broadcast Address (DBA):}$
When a host in a network wants to send some data to all the hosts in some other network.
By seeing we understand that the given DBA belongs to a class C IP address and the default subnet mak of class C ip address is $255.255.255.0$
Now, in DBA's, there will be all $1's$ in Host ID part.
$201.15.16.31 = 201.15.00010000.00011111$
So we can conclude that the last $5$ bits are host bits.
∴ in the subnet mask , there will be $\color{red}{27} \hspace{0.1cm}1's$ Or $\color{red}{28} \hspace{0.1cm} 1's$ Or $\color{red}{29} \hspace{0.1cm} 1's$ Or $\color{red}{30} \hspace{0.1cm} 1's$ Or $\color{red}{31} \hspace{0.1cm} 1's$
The possible subnet masks are :
$\color{red}{27} \hspace{0.1cm}1's$ |
$255.255.255.11100000$ |
$255.255.255.224$ |
Network ID : $201.15.16.0$ |
$\color{red}{28} \hspace{0.1cm}1's$ |
$255.255.255.11110000$ |
$255.255.255.240$ |
Network ID : $201.15.16.16$ |
$\color{red}{29} \hspace{0.1cm}1's$ |
$255.255.255.11111000$ |
$255.255.255.248$ |
Network ID : $201.15.16.24$ |
$\color{red}{30} \hspace{0.1cm}1's$ |
$255.255.255.11111100$ |
$255.255.255.252$ |
Network ID : $201.15.16.28$ |
$\color{red}{31} \hspace{0.1cm}1's$ |
$255.255.255.11111110$ |
$255.255.255.254$ |
Network ID : $201.15.16.30$ |
$\color{blue}{32} \hspace{0.1cm} 1's$ isn't possible , as when the subnet is having $\color{blue}{32} \hspace{0.1cm} 1's$, then the subnet will be $255.255.255.255$, but DBA is $201.15.16.31$, so subnetting must be done & the subnet will be having at most $\color{red}{31} \hspace{0.1cm} 1's$ and at least $\color{red}{27} \hspace{0.1cm} 1's$ .