Torque equation for rod about O, ${\mathrm{I}}_0\mathrm{\alpha }=Fl-\mathrm{N}\frac{l}{3}$

$\therefore \frac{{\mathrm{ml}}^2}{3}\cdot \mathrm{\alpha }=\mathrm{Fl}-\frac{\mathrm{Nl}}{3}$

$\mathrm{m}\frac{\mathrm{l\alpha }}{3}=\mathrm{F}-\frac{\mathrm{N}}{3}$             …(i)

The motion of the sphere is translational only acceleration of the sphere (a) = acceleration of point P

$\frac{\mathrm{l\alpha }}{3}=\mathrm{a}$                        …(ii)

From (i) and (ii) $\mathrm{ma}=\mathrm{F}-\frac{\mathrm{N}}{3}$            …(iii)

For sphere, $\mathrm{N}=\mathrm{ma}$                          …(iv)

Solving (iii) and (iv) we get, $\text{a=}\frac{3\mathrm{F}}{4\mathrm{m}}\therefore \mathrm{N}=\frac{3\mathrm{F}}{4}$

Acceleration of the COM of the rod is

${\mathrm{a}}_{\mathrm{cm}}=\mathrm{\alpha }\frac{\mathrm{l}}{2}=\frac{3\mathrm{a}}{2}=\frac{9\mathrm{F}}{8\mathrm{m}}$

Let horizontal component of hinge reaction is ${\mathrm{N}}_{\mathrm{x}}$. Force equation for rod, 

$\mathrm{F}-\frac{3\mathrm{F}}{4}+{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{c}}\Rightarrow \mathrm{F}-\frac{3\mathrm{F}}{4}+{\mathrm{F}}_{\mathrm{x}}=\mathrm{m}\left(\frac{9\mathrm{F}}{8\mathrm{m}}\right)$

${\mathrm{F}}_{\mathrm{x}}=\frac{9\mathrm{F}}{8}-\frac{\mathrm{F}}{4}=\frac{7\mathrm{F}}{8}=\frac{7\times 80}{8}=70 \mathrm{N}$