Answer
$0.375\;\rm J$
Work Step by Step
Since the can is filled with uniform dense food, its center of mass is at its geometric center.
Now we have two motions of the can while rolling on the floor, the translational motion ($v_{cm}=1$ m/s) and the rotational motion.
This means that the can's kinetic energy is given by
$$K_{tot}=K_{translational }+K_{rotational}$$
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I\omega^2$$
where $\omega$ is the can's angular velocity.
We know that the linear speed is given by $v=R\omega$, Hence $\omega=v/R$.
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I\left[\dfrac{v_{cm}}{R}\right]^2$$
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I \dfrac{v_{cm}^2}{R^2} $$
Recall that the moment of inertia of a disk or a cylinder is $I=\frac{1}{2}mR^2$, and the can is a solid cylinder.
Thus,
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{2}\left(\frac{1}{2}mR^2\right)\dfrac{v_{cm}^2}{R^2} $$
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{4} m \color{red}{\bf\not}R^2 \dfrac{v_{cm}^2}{ \color{red}{\bf\not}R^2} $$
$$K_{tot}=\frac{1}{2}mv_{cm}^2+\frac{1}{4} mv_{cm}^2= \frac{3}{4} mv_{cm}^2 $$
Plugging the given;
$$K_{tot}= \frac{3}{4} (0.500)(1.0^2)=\color{red}{\bf 0.375}\;\rm J $$
