A body of mass $15\ kg$ moving with a velocity of $10\ ms^{-1}$ is bought to rest. The work done by the brake is

$W = change\ in\ kinetic\ energy =\dfrac{1}{2}m(v^2-u^2)$

$W = \dfrac{1}{2} \times 15 \times (0^2-10^2)$

$W = - 750J$

Hence, option $C$ is the correct answer.