= Solution
Part (a) gives
$$
[K_i,K_j]=-ie\hbar\varepsilon_{ijk}B_k.
$$
For
$$
M=\frac i\hbar r\cdot K,
\qquad
N=\frac i\hbar r'\cdot K,
$$
their commutator is the scalar
$$
[M,N]
=\frac{ie}{\hbar}(r\times r')\cdot B.
$$
It therefore commutes with both $M$ and $N$. Applying the stated Baker--Campbell--Hausdorff identity in both orders yields the <magnetic translation algebra>
$$
\boxed{
\mathcal T_r\mathcal T_{r'}
=\exp\left[
\frac{ie}{\hbar}(r\times r')\cdot B
\right]
\mathcal T_{r'}\mathcal T_r}.
$$
Writing $\Phi=B\cdot(r\times r')$, the translations commute exactly when their phase is one:
$$
\frac{e\Phi}{\hbar}=2\pi n,
\qquad n\in\mathbb Z.
$$
Thus the cell flux must be an integral multiple of the <magnetic flux quantum>:
$$
\boxed{\Phi=\frac{2\pi\hbar}{e}n=\frac he n}.
$$
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