 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

At <resonance>, the <Hamiltonian operator> becomes $\hbar V\sigma_x/2$. Therefore
$$
U(\tau)=e^{-iV\tau\sigma_x/2}=R_x(V\tau),\qquad
\boxed{\delta=0,\qquad \tau=\frac{\pi}{2V}.}
$$
It gives the requested positive <rotation about the x-axis>. In the ordered <basis> $(|1\rangle,|0\rangle)$, $|1\rangle$ becomes $(|1\rangle-i|0\rangle)/\sqrt2$, whose <Bloch vector> is $(0,-1,0)$.

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