= Solution
Let $\widehat{\mathbf r}$ point transversely away from the line and define $\widehat{\boldsymbol\varphi}=\widehat{\mathbf x}\times\widehat{\mathbf r}$. Cylindrical <Gauss's law> in the line's <rest frame> gives
$$
\mathbf E'=\frac{\lambda'}{2\pi\epsilon_0r'}\widehat{\mathbf r},\qquad \mathbf B'=0.
$$
The transverse distance is unchanged by the axial boost: $r=r'$. Applying the inverse of the given field transformation to a <boosted uniform line charge>,
$$
\boxed{\mathbf E=\frac{\gamma\lambda'}{2\pi\epsilon_0r}\widehat{\mathbf r},\qquad
\mathbf B=\frac{\mathbf u\times\mathbf E}{c^2}
=\frac{\mu_0\gamma\lambda'u}{2\pi r}\widehat{\boldsymbol\varphi}.}
$$
Here $\gamma=(1-u^2/c^2)^{-1/2}$ and all longitudinal field components vanish. Substituting back verifies $\mathbf B'=\gamma(\mathbf B-\mathbf u\times\mathbf E/c^2)=0$ and $\mathbf E'=\gamma(1-u^2/c^2)\mathbf E=\mathbf E/\gamma$. The sign of $\lambda'$ determines the radial and circulation directions.
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