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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:

$$(2^{2^{B-1}}-1)^2=(2^{2^{B-1}})^2-2*2^{2^{B-1}}+1=2^{2^B}-2^{2^{B-1}+1}+1$$

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 $$\bf o$$ until it is less than $$2^{2^B}$$, then apply $$\bf L$$ until $$2^{2^{B-1}}\le n\lt 2^{2^{B}}$$, and 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 $$10^{10^{58}}$$.