= Solution
The $+x$ <eigenstate> is $|+x\rangle=(|\uparrow\rangle+|\downarrow\rangle)/\sqrt2$, since $\sigma_x(1,1)^T=(1,1)^T$. Use a resonant quarter-turn pulse, $\tau_{\pi/2}=\pi/(2\Omega)$, and choose its <phase> so that
$$
\varphi+\omega_0\tau_{\pi/2}=\frac\pi2\pmod{2\pi}.
$$
The laboratory-frame relative amplitude then is $i e^{-i\pi/2}=1$, giving
$$
\boxed{|\psi(\tau_{\pi/2})\rangle\sim|+x\rangle,\qquad
\langle s_x\rangle=\hbar/2.}
$$
In rotating axes the equivalent target is obtained with $\varphi=\pi/2$; the extra laboratory <phase> above compensates <Larmor precession> during the pulse. After switching the RF field off, the <spin> continues to precess in the static field, so this statement specifies the preparation instant rather than a state fixed along laboratory $x$ forever.
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