Energy required to place the lab in first orbit is the change in potential energy$$+{ k }_{ 0 }{ E }_{ 0 }$$
$$\triangle PE=\cfrac { -GMm }{ 2R } +\cfrac { GMm }{ 2R } =\cfrac { GMm }{ 2R } =\triangle u$$
Velocity given to lab is provided by gravitation force. So, centripetal force=gravitational force
$$\cfrac { m{ v }^{ 2 } }{ 2R } =\cfrac { GMm }{ { \left( 2R \right) }^{ 2 } } \Rightarrow v=\sqrt { \cfrac { GM }{ 2R } } $$
$$KE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { GM }{ 2R } =\cfrac { 1 }{ 4 } \cfrac { GMm }{ R } $$
Total energy$$=\triangle u+{ K.E }_{ 0 }$$
$$=\cfrac { GMm }{ 2R } +\cfrac { 1 }{ 4 } \cfrac { GMm }{ R } $$
$$=\cfrac { 3 }{ 4 } \cfrac { GMm }{ R } $$$$g=\cfrac { GMm }{ { R }^{ 2 } } $$
$$T.E.=\cfrac { 3 }{ 4 } mgR$$
Similarly to shift to orbit of $$3R$$
$$\cfrac { m{ v }^{ 2 } }{ 3R } =\cfrac { GMm }{ { \left( 3R \right) }^{ 2 } } \Rightarrow { v }^{ 2 }=\cfrac { GM }{ 3R } $$
Total energy$$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } m{ v }^{ 2 }$$
$$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } \cfrac { GMm }{ 3R } $$
$$=\cfrac { GMm }{ R } \left( 1-\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } \right) $$
$$=\cfrac { 5 }{ 6 } mgR$$
Energy required$$=\cfrac { 5 }{ 6 } mgR-\cfrac { 3 }{ 4 } mgR$$
$$=\cfrac { \left( 20-18 \right) mgR }{ 24 } $$
Minimum energy required$$=\cfrac { mgR }{ 12 } $$