Let us calculate field at Q in case the cavity weren't there, since mass is proportional to { R }^{ 3 } Gravitational field { g }_{ 0 }=GM(\\dfrac { { 2 }^{ 3 } }{ { 4 }^{ 3 } } )\\dfrac { 1 }{ { 2 }^{ 2 } } =\\dfrac { GM }{ 32 } Also, field due to mass previously in right cavity will be 0 since the point is at its centre. Gravitational field due to left sphere, g'=GM(\\dfrac { { 1 }^{ 3 } }{ { 4 }^{ 3 } } )\\dfrac { 1 }{ { 4 }^{ 2 } } =\\dfrac { GM }{ 1024 } Thus, actual magnetic field at Q is g={ g }_{ 0 }-g'=GM(\\dfrac { 1 }{ 32 } -\\dfrac { 1 }{ 1024 } )=\\dfrac { 31GM }{ 1024 } Answer is option A.