I'll assume the Vigenere cipher is defined to encrypt per $c_i\gets m_i\boxplus k_{(i\bmod l)}$ where $c_i$ is a ciphertext letter, $m_i$ is a plaintext letter, $k_i$ is a key letter, $l$ is the length of the key, $\boxplus$ is the Vigenere table of dimension 26×26. I'll note $\hat k_i=k_{(i\bmod l)}$.
It is assumed
$$\begin{array}{rlcrl}
\Pr[l=1]&=\frac1 2&\quad&\Pr[l=2]&=\frac1 2\\
\Pr[m_0=\mathtt a\;\wedge\;m_1=\mathtt a]&=\frac2 5&\quad&\Pr[m_0=\mathtt a\;\wedge\;m_1=\mathtt b]&=\frac3 5
\end{array}$$
It is asked $p=\Pr[c_0=\mathtt b\;\wedge\;c_1=\mathtt b]$. That is
$$p=\frac2 5\Pr[\hat k_0=\mathtt b\;\wedge\;\hat k_1=\mathtt b]+\frac3 5\Pr[\hat k_0=\mathtt b\;\wedge\;\hat k_1=\mathtt a]$$
Unfortunately, the problem statement is unclear about the distribution of key. We can reasonably assume that each of the possible $k_0$ has probability $\frac1 {26}$, but what about $k_1$ when $l=2$? It is reasonable to assume one of
- Each of the possible $k_1$ has probability $\frac1{26}$. That yields
$$p=\frac2 5\left(\frac1 2\frac1{26}+\frac1 2\frac1{26}\frac1{26}\right)+\frac3 5\left(0+\frac1 2\frac1{26}\frac1{26}\right)=\frac{57}{6760}\approx0.0084$$
(rounding to the nearest, that is down)
- Each of the possible $k_1\ne k_0$ has probability $\frac1{25}$. That yields
$$p=\frac2 5\left(\frac1 2\frac1{26}+0\right)+\frac3 5\left(0+\frac1 2\frac1{26}\frac1{25}\right)=\frac{53}{6500}\approx0.0082$$
(rounding to the nearest, that is up)
Note: Bet on 1 to maximize the probability of scoring a correct answer, even though 2 is the less unsafe cryptographically.
The question's attempted resolution has reading 1, but each of the probability that it sums is missing the multiplication by the probability that $l$ has the value considered.
