Looks like your observation sequence is B,B.
Let's denote the observation at time $t$ as $z_t$ and hidden state at time $t$ as $x_t$. If we denote $\alpha_t(i)$ as the forward values and $\beta_t(i)$ as the backward values, ($i$ is one of the possible hidden states)
$\alpha_t(i)=P(x_t=i,z_{1:t})$
This means $\alpha_t(i)$ is the probability of arriving to state $i$ at time $t$ emitting the observations up to time $t$.
Then,
$\beta_t(i) = P(z_{t+1:T}\mid x_t=i)$ which is the probability of emitting the remaining sequence from $t+1$ until the end of time after being at hidden state $i$ at time $t$.
To do the recursion on $\beta_t(i)$ we can write,
$P(z_{t+1:T}\mid x_t=i)=\sum\limits_jP(x_{t+1}=j,z_{t+1:T}\mid x_{t}=i)$
Using chain rule,
$P(x_{t+1}=j,z_{t+1:T}\mid x_{t}=i) = P(z_{t+2:T},z_{t+1},x_{t+1}=j\mid x_{t}=i)\\
=P(z_{t+2:T}\mid z_{t+1},x_{t+1}=j, x_{t}=i)P(z_{t+1}\mid x_{t+1}=j, x_{t}=i)P(x_{t+1}=j\mid x_{t}=i)$
From conditional independencies of HMM the above probabilities simplifies to
$P(z_{t+2:T}\mid x_{t+1}=j)P(z_{t+1}\mid x_{t+1}=j)P(x_{t+1}=j\mid x_{t}=i)$
Note that $P(z_{t+2:T}\mid x_{t+1}=j) =\beta_{t+1}(j) $ from our definition.
Substituting to $P(z_{t+1:T}\mid x_t=i)$ we get,
$\beta_t(i) = P(z_{t+1:T}\mid x_t=i) = \sum\limits_j \beta_{t+1}(j)P(z_{t+1}\mid x_{t+1}=j)P(x_{t+1}=j\mid x_t=i)$
Now you have a recursion for beta. Last two terms of the last equation you know from your model. Here starting from end of the chain (T) we go backward calculating all $\beta_t$, hence the backward algorithm. In forward you have to start from the beginning and you go to the end of chain.
In your model you have to initialize $\beta_T(i) = P(\emptyset \mid x_T=i)=1$ for all $i$.
This is the probability of not emitting observations after $T=2$.
