Water in a sphere..

When water is liquid in the sphere it will have no rotational kinetic energy also ${m}_{sphere}={m}_{Water}$ , so we write kinetic energy of the system as :

${K.E}_{1} = \frac { 1 }{ 2 } { m }_{ sphere }{ v }^{ 2 }+\frac { 1 }{ 2 } { m }_{ water }{ v }^{ 2 }+\frac { 1 }{ 2 } (\frac { 2 }{ 3 } { m }_{ sphere }{ r }^{ 2 }{ \omega }^{ 2 })=\frac { 4 }{ 3 } m{ v }^{ 2 }$

Note I am using the conditions of pure rolling that is $v=\omega r$

When water is solid it will also have rotational kinetic energy. So now we write it's kinetic energy as :

${K.E.}_{2}=\frac { 1 }{ 2 } { m }_{ sphere }{ v }^{ 2 }+\frac { 1 }{ 2 } { m }_{ water }{ v }^{ 2 }+\frac { 1 }{ 2 } (\frac { 2 }{ 3 } { m }_{ sphere }{ \omega }^{ 2 }{ r }^{ 2 })+\frac { 1 }{ 2 } (\frac { 2 }{ 5 } { m }_{ Water }{ \omega }^{ 2 }{ r }^{ 2 })\\ =\frac { 23 }{ 15 } m{ v }^{ 2 }$

Dividing these two we have :

$\boxed{\frac { { K.E }_{ 1 } }{ { K.E }_{ 2 } } =\frac { 20 }{ 23 }}$

Pratik you should mention in the question that we have to assume same velocity in both the cases.