Since there seems to have been some doubt about whether the electric field of a boosted magnetic dipole is nonzero, this paper by Hnizdo (what a great name!) may be helpful. Section 3 explicitly calculates the field. This is all tangentially related to the Mansuripur paradox as well.
I think dj_mummy's technique works, but it has the disadvantage that it requires you to do a calculation for some finite $l$ and the take the limit $l\rightarrow0$. Here's a different technique, which avoids that.
The vector potential due to a magnetic dipole at rest is $A^\mu=(\phi,\textbf{A})$, with $\phi=0$ and $\textbf{A}=\textbf{m}\times\hat{\textbf{r}}/r^2$. Do a Lorentz boost on this vector, and you get a potential $A^{\mu'}$ for a moving dipole. In fact, you only care about the timelike component of this, so you don't need to compute the rest. Integrate this over all the dipoles (each with its own position and velocity vector), and then take the gradient to get the electric field. Note that although the electric field is $-\nabla\phi-\partial\textbf{A}/\partial t$, the second term can be neglected; the integrated value of $\textbf{A}$ is constant in the $\mu'$ frame, since the dipole current is static in that frame.
Another way to approach this, suggested by Art Brown in a comment below, goes like this. Hnizdo shows that to a sufficient approximation, and ignoring some subtleties related to the definition of multipoles, we can take a magnetic dipole $\textbf{m}$ to have electrical properties, in the lab frame, characterized by an electric dipole moment $\textbf{p}'=\textbf{v}\times \textbf{m}$ (in units with $c=1$). This makes the whole problem look directly analogous to the idea of developing the Biot-Savart law by assembling a collection of magnetic dipoles, so although I haven't worked it out in detail, it sounds like you can get something that's an exact analog of the Biot-Savart law.
The Biot Savart law is
$${\bf B} = \frac{\mu_0}{4\pi} \oint \frac{ I\, d{\bf l} \times {\bf r}}{|{\bf r}|^3}$$
In this case $d{\bf l} \times {\bf r} = dl\,|r|$ directed along the loop axis and integrating around the closed loop leads to a B-field magnitude $ B = \mu_0\, I/2R$ as you suggest.
However, I think there is a problem with your application of Ampere's law. This is that
$$ \oint {\bf B} \cdot d{\bf l} = \mu_0 I\, ,$$
where I is the current enclosed by the closed loop around which you do the line integral on the LHS.
Usually, to apply Ampere's law, you choose a loop to integrate over that has either a constant B-field, and/or with a direction that is either parallel or perpendicular to $d{\bf l}$ (so that the scalar product and/or line integral are much simplified). What loop have you done your integral around? Is the B-field constant along this path? I don't think so...
Best Answer
When the current is time dependent, the time retarded current has to be put into the Biot-Savart integral. If the current is in along straight wire so that you can use cylindrical symmetry, then you can use Ampere's law which does not need the retarded time. Then B=I muo/2pi r For any time dependence in I. In any other case, the Biot-Savart law with the retarded time is very complicated. If the time dependence has a single frequency, as in your case, you can treat the Fourier transform of I (t), but this also is a bit complicated. It is done in advanced EM textbooks.