First of all, notice that by repeatedly applying $\bf o$, we can always reduce a number to 2. This should be readily apparent, but here's a non-rigorous proof of the fact:

2 has a binary representation requiring 2 digits (10), as does 3 (11). The length of the binary representation of a number grows slower than $n$, so $n\ge 3$ implies that $n \buildrel\bf o\over \longrightarrow m$ with $n \gt m$. Therefore, if $n\ge 3$, applying $\bf o$ will always reduce it by at least 1, allowing us through repeated applications to reduce the number to 2.

Now let's consider $\bf o$ in reverse. In order to get to B, we need to apply $\bf o$ to a number whose binary representation has B digits: $2^{B-1} \le n \lt 2^B$. If we square a number, it can jump that range - for example if $B=5$ then $2^{5-1}=16$ and $2^{5}=32$, but $15^2=225$.

However, if we apply $\bf o$ twice, then the range becomes far wider - $2^{2^{B-1}}\le n\lt 2^{2^{B}}$. At this point, there's no way to jump the range:


Because $B\ge 2$, $2^{2^{B-1}+1}\ge 2^{2^{1}+1}=2^3=8$, so $2^{2^B}-2^{2^{B-1}+1}+1<2^{2^B}-7<2^{2^B}$.

So to summarize, apply $\bf L$ until $2^{2^{B-1}}\le n\lt 2^{2^{B}}$, then apply $\bf o$ twice. Of course, this is horribly inefficient, but it proves that it is always possible. For example, with B=197, $2^{2^B}$ is somewhere around $10^{10^{58}}$.