Actually the given question incorporates the concept of horizontal μprogramming (also known as decoded form of control signals) and vertical μprogramming (also known as encoded form of control signals)

The $(a)$ part says :

none or one of the three micro operations of one kind

This is referred to encoding form of vertical one since at most one signal can be active in vertical microprogramming since it involves use of external decoder to select one control signal out of the given control signals..

No of bits required for vertical microprogramming given n number of control signals $=\lceil ( \log_{2} n )\rceil$

Here, $n = 3$

So, no of bits required for part $(a)$ $=\lceil( \log_{2} 3)\rceil= 2$

Now coming to $(b)$ part , it says :

none or upto six micro operations of another kind

at maximum we can have at most $6$ microoperations of another kind at a time. To accommodate that we need decoded form of control signals which is horizontal signals.

So, no of bits required for $(b)$ part

$=$ No of control signals of $(b)$ kind $= 6$

Therefore overall bits required to accommodate both $(a)$ and $(b),$

$ = 2 + 6=8-bits$

Besides this, address field, flags etc are also there in a control word. That is why it is asked in the question :

minimum number of bits in the micro-instruction required

Hence, minimum no of bits required $= 8-bits$

C is the correct answer.