Since the function is onto from a $20$-element set to a $20$-element set, it must be a bijection.

So we are really counting permutations $f$ of $\{1,2,\dots,20\}$ satisfying:

$f(k) \text{ is a multiple of } 3 \quad \text{whenever } k \text{ is a multiple of } 4.$

First identify the domain elements that are multiples of $4$:

$\{4,8,12,16,20\}$

There are $5$ such elements.

Their images must be multiples of $3$ in $\{1,2,\dots,20\}$:

$\{3,6,9,12,15,18\}$

There are $6$ such elements.

Because $f$ is a bijection, the images of the $5$ special inputs must be distinct. So we must assign to

$\{4,8,12,16,20\}$

any $5$ distinct values chosen from the $6$ multiples of $3$.

Number of ways:

$^6P_5 = \frac{6!}{(6-5)!} = 6!$

After that, one multiple of $3$ remains unused, and also the other $15$ domain elements remain:

$20-5=15$

These $15$ remaining domain elements can be mapped bijectively to the $15$ remaining codomain elements in:

$15!$

ways.

Hence total number of such onto functions is

$6!\cdot 15!$

Compare with the options:

• A: $65\times 15!$
• B: $56\times 15$
• C: $(15)!\times 6!$
• D: $5!\times 6!$

Therefore the correct option is:

$\boxed{\text{C}}$