Weight (\(W\)) = \(Mg\)

Initially for the equilibrium \(F + F = Mg\)   

\(F = Mg/2\)

When one man withdraws, the torque on the rod

\(\tau  = I\alpha  = Mg\frac{l}{2}\)

 \(\frac{{M{l^2}}}{3}\alpha  = Mg\,\frac{l}{2}\)                        [As \(I = Ml^2/ 3\)]

Angular acceleration \(\alpha  = \frac{3}{2}\frac{g}{l}\)and linear acceleration \(a = \frac{l}{2}\alpha  = \frac{{3g}}{4}\)

Now if the new normal force at \(A\) is \(F'\) then \(Mg - F' = Ma\)

 \(F' = Mg - Ma = Mg - \frac{{3Mg}}{4}\)\( = \frac{{Mg}}{4}\)\( = \frac{W}{4}\).