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Let ܵ$S$ denote the set of all functions $f:\{0,1\}^4 \to \{0,1\}$. Denote by $N$ the number of functions from S to the set $\{0,1\}$. The value of $\log_2 \log_2N$ is _______.

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For a function from set $A$ to set $B$, we need to have a mapping for all elements of $A$ and mapping must be unique.
Let number of elements in $A$ be $m$ and that in $B$ be $n$

So, if we consider an element from $A,$ it can be mapped to any of the element from $B.$ i.e., it has $n$ possibilities when a function is formed. Similarly, for all other members also there are $n$ possibilities as one element from $A$ can be mapped to  only a single element in $B$ (though reverse need not true). So, for $n$ elements in $A,$ we totally have $\underbrace{n \times \dots \times n}_{m \text{ times}} = n^m$ possible functions.

In the question Number of elements (functions) in $f$ is $2^{2^4}$ as $\{0,1\}^4$ contains $2^4$ elements. So, number of functions from $S$ to $\{0,1\}$ will be $2^{2^{2^4}}$. So, $\log_2 \log_2 N = 2^4 = 16.$
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How this 2^2^4 came ?? Also what does {0,1}4   Denotes? Please explain

+21

{0, 1}4 = {0000, 0001, 0010, .... , 1111} set of all 4 length strings over {0, 1}. So, 16 elements in it.

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what does {0,1}^4 mean?
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4 places, each place can be filled with 0 or 1 , total possible combinations, 2 choices at each places, 2*2*2*2, 4-bit binary numbers,0000 to 1111.

Number of functions in S is 2^16 so in N is (2^(2^16)), and value of  loglog n will be 16.

f:{0,1}4→{0,1}

{0,1}4 contains total 24 elements

so 16 elements and S will be [co-domain]domain  = 216

N is S to {0,1} so 2^216

log2log2N  = 16

If anybody wondering why it is 2^2^16 and not just 2^16, here is explanation.

From above two answer number of functions are 2^16 .{f1 , f2 , f3 , f4, ...............f2^16}

And if you read question carefully they asking number of functions from S to the set {0,1}

and set S is .{f1 , f2 , f3 , f4, ...............f2^16} which has to mapped to set {0, 1}

I have written this answer according to my understanding , please correct me if i am wrong