The position of center the combined system with B as origin.

$\mathrm{y}=\frac{0.6\times 30+0.3\times 45}{0.6+0.3}=35\mathrm{cm}\text{ above the table }$

If we take "rod + particle" as system we have no external force acting on the system in horizontal direction hence we can conserve the linear momentum of the system in horizontal direction just before and just after collision.

Let ${\mathrm{v}}_{\mathrm{CM}}$ be the linear velocity of CM and $\mathrm{\omega }$ is the angular velocity about an axis passing through CM immediately after collision. From conservation of linear momentum, we get

${\mathrm{mv}}_0=(\mathrm{M}+\mathrm{m}){\mathrm{v}}_{\mathrm{CM}}\text{ or }{\mathrm{v}}_{\mathrm{CM}}=\frac{{\mathrm{mv}}_0}{\mathrm{M}+\mathrm{m}}=\frac{(0.3)\left(24\right)}{0.6+0.3}=8\mathrm{m}/\mathrm{s}$

The system has no external impulse acting on the system. Hence we can conserve angular momentum of the system about any point. Let us conserve angular momentum about C.M of the system.

Angular momentum of the system just before collision. 

${\overrightarrow{\mathrm{L}}}_{\text{initial }}={\mathrm{mv}}_0(\mathrm{h}-\mathrm{y})(-\hat{\mathrm{k}})=0.3\times 24\times (0.35-0.30)(-\hat{\mathrm{k}})$

Just after collision ${\overrightarrow{\mathrm{L}}}_{\text{initial }}={\mathrm{I}}_{\mathrm{cm}}\overrightarrow{\mathrm{\omega }}$

${\mathrm{I}}_{\mathrm{cm}}={\left({\mathrm{I}}_{\mathrm{cm}}\right)}_{\mathrm{rod}}+{\left({\mathrm{I}}_{\mathrm{cm}}\right)}_{\text{particle }}=\frac{{\mathrm{Ml}}^2}{12}+\mathrm{M}{\left(\mathrm{y}-\frac{\mathrm{l}}{2}\right)}^2+\mathrm{m}(\mathrm{h}-\mathrm{y}{)}^2$

$\Rightarrow {\mathrm{I}}_{\mathrm{cm}}=\frac{(0.6)(0.6{)}^2}{12}+0.6(0.35-0.30{)}^2+0.3(0.45-0.35{)}^2$

$=0.0225{\mathrm{kgm}}^2$

Hence ${\overrightarrow{\mathrm{L}}}_{\mathrm{final}}=0.0225\mathrm{\omega }(-\hat{\mathrm{k}})$

As ${\overrightarrow{\mathrm{L}}}_{\text{initial }}={\overrightarrow{\mathrm{L}}}_{\text{final }}\Rightarrow 0.75(-\hat{\mathrm{k}})=0.0225\mathrm{\omega }(-\hat{\mathrm{k}})$

Or $\mathrm{\omega }=\frac{0.72}{0.225}=32\mathrm{rad}/\mathrm{s}$

Initially the rod was in vertical orientation then it become horizontal. It means angle rotated by the rod is $\frac{\mathrm{\pi }}{2}$.

Hence time taken in this process, $\mathrm{t}=\frac{\mathrm{\pi }/2}{\mathrm{\omega }}=\frac{\mathrm{\pi }}{2\times 32}=\frac{\mathrm{\pi }}{64}\sec$