Answer
$P=1.7\times10^{25}W$.
Work Step by Step
The pulsar’s power output is its daily energy loss, divided by the elapsed time of 86400s.
$$P=\frac{\Delta KE}{\Delta t}$$
The rotational kinetic energy is $\frac{1}{2}I\omega^2$.
$$P=\frac{\frac{1}{2}I\omega^2(fraction\;lost)}{\Delta t}$$
$$P=\frac{\frac{1}{2}\frac{2}{5}MR^2\omega^2(fraction\;lost)}{\Delta t}$$
$$P=\frac{1}{5}\frac{(1.5)(1.99\times10^{30}kg)(8000m)^2(2\pi\;rad/s)^2(1\times10^{-9})}{86400s}$$
$$P=1.7\times10^{25}W$$
