The system can be described by a 5 states Markov chain, with the state described by a number of consecutive six's accumulated so far. The transition matrix is:
$$
P = \begin{bmatrix}
1-p & p & 0 & 0 & 0 \\
1-p & 0 & p & 0 & 0 \\
1-p & 0 & 0 & p & 0 \\
1-p & 0 & 0 & 0 & p \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
$$
where for the case at hand $p=\frac{1}{6}$ is the probability to get a six at the next rolling of the die.
The classic way to solve this is to consider the expected number of rolls $k_i$ given an initial state $i$. We are interested in computing $k_0 = \mathbb{E}(X)$.
Conditioning on a first move, the following recurrence equation holds true:
$$
k_i = 1 + p k_{i+1} + (1-p) k_0 \qquad \text{for} \qquad i=0,1,2,3
$$
with boundary condition $k_4 = 0$. Solving this linear system yields:
$$
k_0 = \frac{p^3+p^2+p+1}{p^4} \quad
k_1 = \frac{p^2+p+1}{p^4} \quad
k_2 = \frac{p+1}{p^4} \quad
k_3 = \frac{1}{p^4} \quad
k_4 = 0
$$
In[24]:= Solve[{k0 == 1 + k1 p + (1 - p) k0,
k1 == 1 + k2 p + (1 - p) k0, k2 == 1 + p k3 + (1 - p) k0,
k3 == 1 + p k4 + (1 - p) k0, k4 == 0}, {k1, k2, k3, k0,
k4}] // Simplify
Out[24]= {{k1 -> (1 + p + p^2)/p^4, k2 -> (1 + p)/p^4, k3 -> 1/p^4,
k0 -> (1 + p + p^2 + p^3)/p^4, k4 -> 0}}
Substituting $p=\frac{1}{6}$ gives $k_0 = \mathbb{E}(X) = 1554$.
Added A good reference on the subject is a book by J.R. Norris, "Markov chain" (
Amazon). The chapter on discrete Markov chains is
available on-line for free from the author. Section 1.3 discusses finding mean hitting times $k_i$.