Solution

first Analyse the rest frame S'

We have a long cylindrical solenoid, with axis along z'.
rest magnetic moment
$\mathbf{M}' = M\,\hat{\mathbf{z}}'$
There is no electric dipole moment: $\mathbf{p}' = 0$

In the laboratory system S, the solenoid moves with velocity
$\mathbf{v} = v\,\hat{\mathbf{x}}$

This is the same situation as in the problem 14.3.24 but with the round solenoid

Under a boost with velocity $\mathbf{v}$
the electric and magnetic dipole moments$\mathbf{p}$and $\mathbf{m}$(in CGS) transform as

\begin{aligned} \mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\ \mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right). \end{aligned}

With of course

$\mathbf{p}' = 0$ $\mathbf{v}\cdot\mathbf{m}' = 0$

so this simplifies to

$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad \mathbf{m} = \gamma\,\mathbf{M}.$

Expression in SI

And finaly In the International System, the equivalent transformation is

$\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad \mathbf{m} = \gamma\,\mathbf{M}$