The easiest way to tackle the problem is probably to treat the series of trials as a $4$-state Markov chain, in which the states are outcomes of the previous two tests if these have been $00$, $01$ or $10$, or the state $F$ of failure. Numbering these states $\ 1,2,3,4\ $, respectively, the chain starts in state $1$, and the transition matrix is
$$
P=\begin{pmatrix}
\frac{2}{3}&\frac{1}{3}&0&0\\
0&0&\frac{2}{3}&\frac{1}{3}\\
\frac{2}{3}&0&0&\frac{1}{3}\\
0&0&0&1
\end{pmatrix}\ .
$$
The probability that the chain is in state $\ j\ $ after $\ y\ $ tests is $\ \delta_1^TP^y\delta_j\ $, where $\ \delta_j\ $ is the $\ j^\mathrm{\,th}\ $ column of the $\ 4\times 4\ $ identity matrix. The probability that $\ Y \le y\ $ is just the probability that the chain is in state $\ 4\ $ after $\ y\ $ tests—that is, $\ \delta_1^TP^y\delta_4\ $. Thus, the probability that $\ Y=y\ $ is $\ \delta_1^TP^y\delta_4-\delta_1^TP^{y-1}\delta_4=\delta_1^TP^{y-1}\left(P-I\right)\delta_4\ $. If $\ z_y=\delta_1^TP^{y-1}\left(P-I\right)\delta_4\ $, then, since $\ x^4-\frac{5}{3}x^3+ \frac{2}{3}x^2-\frac{4}{27}x+\frac{4}{27}\ $ is the characteristic polynomial of $\ P\ $ it follows that $\ z_y\ $ satisfies the recursion $\ z_{y+4}=\frac{5}{3}z_{y+3}+ \frac{2}{3}z_{y+2}-\frac{4}{27}z_{y+1}+\frac{4}{27}z_y\ $. Once the first four values of $\ z_y\ $ have been calculated, this recursion enables the subsequent values to be calculated without needing to do any further matrix multiplications.
Here's a table of the values of $\ \mathrm{Prob}\left(Y=y\right)\ $ for $\ y\ $ from $1$ to $20$:
$$
\begin{matrix}
y & \mathrm{Prob}\left(Y=y\right)
\mbox{(exact)} &\mbox{(approx.)}\\
1&0&0\\
2&\frac{1}{9}& 0.11\\
3&\frac{4}{27}& 0.15\\
4&\frac{8}{81}& 0.099\\
5&\frac{20}{243}&0.082\\
6&\frac{56}{3^6}&0.077\\
7&\frac{144}{3^7}&0.066\\
8&\frac{368}{3^8}&0.056\\
9&\frac{960}{3^9}&0.49\\
10&\frac{2,946}{3^{10}}&0.042\\
11&\frac{6,464}{3^{11}}&0.036\\
12&\frac{16,768}{3^{12}}&0.032\\
13&\frac{43,520}{3^{13}}&0.027\\
14&\frac{112,896}{3^{14}}&0.023\\
15&\frac{292,864}{3^{15}}&0.020\\
16&\frac{759,808}{3^{16}}&0.018\\
17&\frac{1,971,200}{3^{17}}&0.015\\
18&\frac{5,113,856}{3^{18}}&0.013\\
19&\frac{13,266,944}{3^{19}}&0.011\\
20&\frac{34,418,688}{3^{20}}&0.0099\\
\end{matrix}
$$
If $\ e_1, e_2, e_3\ $ are the expected number of tests from states $\ 1,2,3\ $, respectively, until ultimate failure, then these $\ e_i\ $ must satisfy the equations
\begin{eqnarray}
e_1&=&1 + \frac{2}{3}e_1 + \frac{1}{3}e_2\\
e_2&=&1 + \frac{2}{3}e_3 \\
e_3&=&1 + \frac{2}{3}e_1\ ,
\end{eqnarray}
Which give $\ e_1 = 3 + e_2 = 4+ \frac{2}{3}e_3=\frac{14}{3}+\frac{4}{9}e_1\ $. Thus, $\ \frac{5}{9}e_1=\frac{14}{3}\ $, or $\ e_1=\frac{42}{5}=8.4\ $. This is probably where your hint comes from, but it's a little misleading, because the approximation $\ \sum_\limits{y=1}^{20} y\,\mathrm{Prob}\left(Y=y\right)\ $ is only about $\ 6.7\ $. You need to take $\ n=68\ $ for $\ \sum_\limits{y=1}^n y\,\mathrm{Prob}\left(Y=y\right)\ $ to approximate $8.4$ correctly to two decimal places.
