= Solution
The <divergence theorem> states that a continuously differentiable <vector field> $\mathbf F$ on a volume $V$ with piecewise smooth closed boundary $S$ satisfies
$$
\int_V\nabla\cdot\mathbf F\,dV=\int_S\mathbf F\cdot\hat{\mathbf n}\,dS,
$$
where $\hat{\mathbf n}$ is the outward <unit normal>. Take $\mathbf F=\nabla a\times\mathbf B$. Since the <curl of a gradient> is zero for a twice continuously differentiable <scalar field>, part (i) gives
$$
\nabla\cdot(\nabla a\times\mathbf B)=-\nabla a\cdot(\nabla\times\mathbf B).
$$
Integrating this identity and applying the <divergence theorem> proves
$$
\boxed{\int_V\nabla a\cdot(\nabla\times\mathbf B)\,dV=-\int_S(\nabla a\times\mathbf B)\cdot\hat{\mathbf n}\,dS.}
$$
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