= Solution
Let $M$ be the central stellar mass and take positive <Keplerian rotation>, $\Omega(R)=\sqrt{GM/R^3}$. The initial <poloidal magnetic field> stays fixed by part (a), while $R\Omega'(R)=-(3/2)\Omega(R)$. Integrating the toroidal induction equation with $B_\phi(R,0)=0$ gives
$$
\boxed{B_R(R,t)=B_0\frac{R_0}{R},\qquad B_z(R,t)=0,\qquad B_\phi(R,t)=-\frac32\Omega(R)t\,B_0\frac{R_0}{R}.}
$$
The initial field is a <solenoidal vector field> on the disk region $R>0$, since $\partial_R(RB_R)=0$. Assuming $B_0\ne0$, the amplitude ratio is
$$
\frac{|B_\phi|}{|\mathbf B_p|}=\frac32\Omega t.
$$
It reaches one at
$$
\boxed{t_{\phi=p}=\frac2{3\Omega(R)}=\frac{P_{\rm orb}(R)}{3\pi},\qquad P_{\rm orb}=\frac{2\pi}{\Omega}.}
$$
Thus the equality time is the same fraction $1/(3\pi)\approx0.106$ of the orbital period at every radius. If “orbital time” instead means $\Omega^{-1}$, the corresponding fraction is $2/3$. These are kinematic solutions with prescribed rotation and the neglected magnetic back-reaction.
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