Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A to B such that for every element bin B, there is at least one element a in A such that f(a) = b. Then, the total number of such functions f is

The correct option is 540.

To find the total number of such functions, we need to calculate the number of onto (surjective) functions from set A to set B.
Let us first identify the given sets and their cardinalities:
Set A = {2, 3, 5, 7, 11, 13}, which contains 6 elements. Thus, $\left|A\right|=6$.
Set B = {1, 8, 27}, which contains 3 elements. Thus, $\left|B\right|=3$.

The condition that for every element $b$ in B, there is at least one element $a$ in A such that $f\left(a\right)=b$ means that the function $f$ must be onto (surjective).

We can use the Principle of Inclusion-Exclusion to find the total number of onto functions from a set of size $n$ to a set of size $m$.
Here, $n=6$ and $m=3$.

The formula for the number of onto functions is:
$N=m^n-\left(\genfrac{}{}{0ex}{}{m}{1}\right){(m-1)}^n+\left(\genfrac{}{}{0ex}{}{m}{2}\right){(m-2)}^n-...$

Substituting $n=6$ and $m=3$ into the formula, we get:
$N=3^6-\left(\genfrac{}{}{0ex}{}{3}{1}\right)2^6+\left(\genfrac{}{}{0ex}{}{3}{2}\right)1^6$

Now, let us calculate the value of each term:
1. Total number of unrestricted functions:
$3^6=729$
2. Functions missing at least one element of B:
$\left(\genfrac{}{}{0ex}{}{3}{1}\right)2^6=3\times 64=192$
3. Functions missing at least two elements of B:
$\left(\genfrac{}{}{0ex}{}{3}{2}\right)1^6=3\times 1=3$

Combining these terms using the Principle of Inclusion-Exclusion:
$N=729-192+3$
$N=537+3$
$N=540$

Therefore, the total number of such functions is 540.