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Assume you need a $64$-to-$1$ multiplexer controlled by $6$ select signals $A$ through $F,$ i.e., a binary number $ABCDEF$ specifies the index of the input to be sent to the output. However, as we assume a logic gate can not have more than $8$ inputs, such a MUX is not directly available. So you build this $64$-to-$1$ MUX using $4$-to-$1$ mux, such that a minimum number of $4$-to-$1$ mux are used. We build this $64$-to-$1$ MUX using three levels of $4$-to-$1$ MUXs where level $3$ has one $4$ -to-$1$ mux and its output is the output of the original $64$-to-$1$ mux.
The realization of this $64$-to-$1$ mux using three levels of $4$-to-$1$ muxes is such that Level $i(i=1,2,3)$ is controlled by $X_{i} Y_{i}$ select signals i.e. In level $i,$ the most significant select line of each $4$-to-$1$ mux is connected to $X_{i}$ and the least significant control line of each $4$-to-$1$ mux is connected to $Y_{i}$.
To select input $27$ i.e. $(\mathrm{ABCDEF}=011011)$ and send it to the output, the value of $X_{1} Y_{1} X_{2} Y_{2} X_{3} Y_{3}$ should be

  1. $011011$
  2. $110110$
  3. $111001$
  4. $100111$
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Use three levels of $4\times 1$ MUXs. $16$ in first level controlled by $EF$ to receive $64$ inputs and generate $16$ outputs, $4$ in $2$nd level controlled by $CD$ to receive $16$ outputs from first level and generate $4$ outputs, and one in last level controlled by $AB$ to select one of the four outputs from the $2$nd level as the final output.

So, $X_{1} Y_{1} \equiv E F ; X_{2} Y_{2} \equiv C D ; X_{3} Y_{3} \equiv A B$

Hence, answer is Option C.

NOTE : A common mistake student might do is that they might assign $AB$ to the select lines of first level. This will not give desried behavious of Mux.
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