Massless Kerr black hole Kerr metric has the following form:
$$
ds^2 = -\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 +
       \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 +
    \left(r^2+a^2\cos(\theta)\right) d\theta^2 
  + \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 -
    \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt
$$
This metric describes a rotating black hole.
If one considers $M=0$:
$$
ds^2 = - dt^2 +
       \left(\frac{r^2+a^2\cos^2(\theta)}{r^2+a^2}\right) dr^2 +
    \left(r^2+a^2\cos(\theta)\right) d\theta^2 
  +
    \left(r^2+a^2\right)\sin^2(\theta) d\phi^2 
$$
This metric is a solution of the Einstein equations in vacuum.
What is the  physical interpretation of such a solution?

It's simply flat space in Boyer-Lindquist coordinates. By writing
$\begin{cases}
x=\sqrt{r^2+a^2}\sin\theta\cos\phi\\
y=\sqrt{r^2+a^2}\sin\theta\sin\phi\\
z=r\cos\theta
\end{cases}$
you'll get good ol' $\mathbb{M}^4$.
