The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

$I=I_0{\left[\frac{\sin \alpha }{\alpha }\right]}^2,$  where  $\alpha =\frac{\phi }{2}$
For  $n^{th}$ secondary maxima  $d\sin \theta =\left(\frac{2n+1}{2}\right)\lambda$
$\Rightarrow \alpha =\frac{\phi }{2}=\frac{\pi }{\lambda }\left[d\sin \theta \right]=\left(\frac{2n+1}{2}\right)\pi$
$\therefore I=I_0{\left[\frac{\sin \left(\frac{2n+1}{2}\right)\pi }{\left(\frac{2n+1}{2}\right)\pi }\right]}^2$
So  $I_0:I_1:I_2=I_0:\frac{4}{9{\pi }^2}{I}_0:\frac{4}{25{\pi }^2}{I}_0$
$=1:\frac{4}{9{\pi }^2}:\frac{4}{25{\pi }^2}$