When a solid sphere rolls without slipping on a surface, its rotational kinetic energy is $40J.$ Then its total kinetic energy is

We know that rotational $K.E=\left(\frac{K^2}{K^2+R^2}\right)$ total energy
Total kinetic energy $=\left(\frac{K^2+R^2}{K^2}\right)$ rotational K.E
For sphere $K^2=\frac{2}{5}{R}^2$  
Then, total $K.E=\left(\frac{2/5R^2+R^2}{2/5R^2}\right)\times 40=\frac{7}{2}\times 40=140J$  
So, the correct answer is 3