Let the first term be a and the common ratio be r. The third term is given by:

\(ar^2 = 4a\)

Solving for r, we get:

\(r^2 = 4\)

Thus, \(r = \pm 2\).

The sum of the first six terms of a geometric progression is given by:

\(S_6 = a \frac{r^6 - 1}{r - 1}\)

We know \(S_6 = ka\), so:

\(a \frac{r^6 - 1}{r - 1} = ka\)

Canceling a from both sides, we have:

\(\frac{r^6 - 1}{r - 1} = k\)

Substituting \(r = 2\):

\(\frac{2^6 - 1}{2 - 1} = k\)

\(\frac{64 - 1}{1} = k\)

\(k = 63\)

Substituting \(r = -2\):

\(\frac{(-2)^6 - 1}{-2 - 1} = k\)

\(\frac{64 - 1}{-3} = k\)

\(k = -21\)

Therefore, the possible values of k are 63 and -21.