= Solution
Adopt the boundary convention $a_0=a_n=0$. Since $L$ has nonzero entries only on its first off-diagonals and $B$ only on its second off-diagonals, $BL$ and $LB$ can contribute to $[B,L]$ only on the first and third off-diagonals. On the upper first off-diagonal,
$$
(BL)_{j,j+1}=a_ja_{j+1}^2,
\qquad
(LB)_{j,j+1}=a_{j-1}^2a_j,
$$
so
$$
[B,L]_{j,j+1}=a_j(a_{j+1}^2-a_{j-1}^2).
$$
The lower first off-diagonal is equal to this by part (a). On the third off-diagonal the two products cancel:
$$
(BL)_{j,j+3}=a_ja_{j+1}a_{j+2}=(LB)_{j,j+3}.
$$
The diagonal and every remaining entry vanish. Therefore, if
$$
c_j=a_j(a_{j+1}^2-a_{j-1}^2),
$$
then
$$
\boxed{
[B,L]=
\begin{pmatrix}
0&c_1&0&\
c_1&0&c_2&\
0&c_2&0&\
&&&\ddots
\end{pmatrix}.
}
$$
This is the commutator calculation in the <Zero-diagonal Jacobi Lax pair for the finite Volterra lattice>.
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