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     Let the mass of the disc be M.  Let the mass of the man be  m1.  Let the mass of each block be  m2.    Let the radius of the disc be R.  Le the angular velocity of the disc be  ω.

The moment of inertia of the disc, about the axis passing through its center and perpendicular to its surface.

    I = I= \int\limits^R_0 {r^2*(\rho *2\pi r} \, dr)\\\\=\frac{M}{\pi R^2}*2\pi \int\limits^R_0 {r^3} \, dr\\\\=\frac{2M}{R^2}\frac{R^4}{4}\\\\=\frac{MR^2}{2}

Moment of inertia of the man:    0  as he sits at the center.  So distance of the person from the axis of rotation is 0.

  Moment of inertia of the two blocks =  2 * m2 * R²

  The new inertia :  M R² /2 + 2  m2  R²
               =  ( M/2 + 2 m2)  R²

2 5 2