Ok, assuming "mj" in the question is Megajoules MJ)Â you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6Â J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is
E = (1/2)Iω² Â
where I is the moment of inertia of the fly wheel and ω is the angular velocity. Â
So this comes down to finding the value of I that gives the required energy. Â We know the mass is 101kg. Â The formula for a solid cylinder's moment of inertia is
 I = (1/2)mR²
We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec.  However, ω must be in radians per second so multiply it by 2Ï€ to getÂ
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω²,  for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²
Now find the radius R,
 165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m
R = 1.807m