As man alight, its angular momentum about a point on platform is conserved. 

By angular momentum conservation,

$L_{\mathrm{initial} }=L_{\mathrm{final} }\Rightarrow mvr=I\omega$

$\Rightarrow mvr=\frac{m{l}^2}{3}·\omega \Rightarrow \frac{3vr}{l^2}=\omega$

or $\omega =\frac{3\times 2\times \left(\frac{1.5}{2}\right)}{{\left(1.5\right)}^2}=2 \mathrm{rad}/\mathrm{s}$

So, instantaneous angular speed of man $=2 \mathrm{rad}/\mathrm{s}$.