Solving through Options, I am directly taking option (B) based on that you can also verify why option (A) is wrong.

B ) A-216.85.102.63 ,B-216.85.102.31 ,C- 216.85.102.127

A->$216.85.102.63$

Since requirement of Dept A is 28 so rounding off to power of $2$ which is $2^5 = 32$. So, subnet for 32 IPs will be $255.255.255.224$. Means 3 Subnet bits and 5 host bits.

Given A->$216.85.102.63$ expanding its fourth octet in binary **001**1 1111. Now here bold letters are SN bits and remaining is host bit. Throught its fourth octet it is clear that its is DBA of first subnet. Seems GOOD DBA for our Dept A.

And Address range is $216.85.102.32$ to $216.85.102.63$

Similarly

B->$216.85.102.31$

Since requirement of Dept B is 14 so rounding off to power of $2$ which is $2^4 = 16$. So, subnet for 16 IPs will be $255.255.255.240$. Means 4 Subnet bits and 4 host bits.

Given B->$216.85.102.31$ expanding its fourth octet in binary **0001** 1111. Now here bold letters are SN bits and remaining is host bit. Through its fourth octet it is clear that its is DBA of first subnet. Seems GOOD DBA for our Dept B.

And Address range is $216.85.102.16$ to $216.85.102.31$

Similarly

C->$216.85.102.127$

Since requirement of Dept C is 53 so rounding off to power of $2$ which is $2^6 = 64$. So, subnet for 64 IPs will be $255.255.255.192$. Means 4 Subnet bits and 4 host bits.

Given C->$216.85.102.127$ expanding its fourth octet in binary **01**11 1111. Now here bold letters are SN bits and remaining is host bit. Through its fourth octet it is clear that it is DBA of first subnet. Seems GOOD DBA for our Dept C.

And Address range is $216.85.102.64$ to $216.85.102.127$

Note:- To find Address range make all possible combination from host bit, you will get address range for it.

Hint for Option A) Address ranges are:-

A-> $216.85.102.0$ to $216.85.102.31$

B-> $216.85.102.80$ to $216.85.102.95$

C-> $216.85.102.64$ to $216.85.102.127$

Here you can find that there is** Address Overlapping **between Dept B and Dept C. Hence it can't be the answer.

So, final answer is **Option (B) **