the crab pulsar spins with a period of 33 ms its radius is about 25 km its mass is roughly 4 x 10^30kg (around twice the mass of the sun how much rotational kinetic energy does it have?

1.81*10⁴³J [ or 2*10⁴³J if you care about sigfig rules]

Explanation:

Rotational Kinetic Energy = (1/2) Iω²

I [solid sphere] = (2/5) (mass) (radius) ² ω = period of rotation [in rad/seconds]

Energy of the Crab Nebula = (1/2) (2/5) (4*10³⁰) (25x10³) ² ((2π) / (33*10⁻³)) ² = 1.81*10⁴³J

The formulas can basically all be looked up (or even derived) if you can't remember them. Note that ω is equal to 2π / (period [in seconds]) and not one of its other whacky forms.