The mechanics of this problem are that the chap's actions cannot result in any net external torque on the system and so angular momentum must be conserved. Since his movements must alter the moment of inertia the angular velocity will change. In fact it increases and so the kinetic energy of the system is increased and it is this

**KE GAIN** that is provided for by the work he does! Hence note that KE is NOT conserved!

Overall energy is conserved because the extra KE is provided by the work he does which is provided by the chemical energy stored in his body.

So let $\displaystyle I$ and $\displaystyle \omega$ be the moment of inertia and angular speed respectively when he is at radius $\displaystyle r$ on the disc. Furthermore let $\displaystyle I_0$ and $\displaystyle \omega_0$ correspond to $\displaystyle r=R$ i.e. when he's at the rim.

Then by conservation of angular momentum we have

$\displaystyle I \omega = I_0 \omega_0$

and $\displaystyle I(r) = mr^2 + \frac{1}{2}M R^2$

where $\displaystyle m$ is the mass of the man and $\displaystyle M$ is the mass of the disc.

so we have

$\displaystyle \omega (r) = \frac{I_0}{I(r)} \omega_0$

i.e.

$\displaystyle \color[rgb]{0,0,1} \boxed{\omega (r) = \frac{(M +2 m) R^2}{MR^2 +2 m r^2} \omega_0}$

Now to compute the work done there are two ways.

One way is to consider the system being equivalent to a non-rotating system with gravitational force outwards (centrifugal force) and so the work done is to overcome this force and is hence converted to potential energy (in this frame of reference).

The second way, which is what I'll do, is just work out the KE gained in the system as it doesn't involve any integration.

So

$\displaystyle W = \frac{1}{2}(I_f \omega_f^2 - I_0 \omega_0^2 )$

where the f subscripts denote the final value of the variable when $\displaystyle r=R$.

So using the fact that $\displaystyle I_f \omega_f = I_0 \omega_0$ from conservation of angular momentum we have that the work, $\displaystyle W$, is given by

$\displaystyle W = \frac{1}{2} I_0 \omega_0 \left( \omega_f - \omega_0 \right)$

$\displaystyle = \frac{1}{2} I_0 \omega_0^2 \left( \frac{MR^2 +2 mR^2}{MR^2} -1 \right)$

$\displaystyle = \frac{m}{M} I_0 \omega_0^2$.

So in explicit terms we have

$\displaystyle \color[rgb]{0,0,1} \boxed{W = \frac{m(2m+M)}{2M} R^2 \omega_0^2}$.

I'll leave it to you to put in the numerical values!