A solid sphere of mass 1kg rolls without slipping on a plane...

Let velocity of centre of mass of sphere be $v$.

Total kinetic energy of solid sphere is $KE = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}I{\omega}^{2}$, where, $I$ is moment of inertia of the solid sphere about the axis passing through it's centre of mass and $\omega = \dfrac{v}{R}\left(\text{pure rolling condition}\right)$.

Therefore,

$7 \times {10}^{- 3} = \dfrac{1}{2} \times 1 \times ( v{)}^{2} + \dfrac{1}{2}\left[\dfrac{2}{5} ( 1 ) {R}^{2}\right]{\left(\dfrac{v}{R}\right)}^{2}$

$⇒ \dfrac{1}{2}{v}^{2}\left(\dfrac{7}{5}\right) = 7 \times {10}^{- 3}$

$⇒ {v}^{2} = {10}^{- 2}$

$⇒ v = {10}^{- 1} = 0 . 1m{s}^{- 1} = 10cm{s}^{- 1}$