The total number of blocks is 8192.
Since it is a 4-way set associative cache, each set can contain 4 blocks.
Therefore, total number of sets is $\frac{8192}{4} = 2048 = 2^{11}$.
Hence, number of bits required will be $log_2(2^{11}) = 11$
Also, size of one block = $32 \, \text{words} \Rightarrow 32 \times 64 \, \text{bits} = 2^{11} \, \text{bits}$
Now since the addresses are word addressable, the number of words is $\frac{2^{11}}{2^6} = 2^5$
Therefore, word offset is $log_2(2^{5}) = 5 \, \text{bits} $
Now we know that $$\text{Number of bits in physical address = Tag + Set + Word Offset}$$
Plugging in the values, we get the value of Tag as $48$ bits.
If this is not the correct answer, do let me know. I tried to answer it to the best of my knowledge.