Two spheres of same metal weight …

Since both the spheres are made of same metal, their weights are directly proportional to their volumes, or the ratio of their volumes is equal to the ratio of their weights.
Radius of first small sphere, r₁ = 3 cm.
Let the radius of second small sphere be r₂ cm.
$\therefore \quad \frac { \text { volume of first small sphere } } { \text { volume of second small sphere } } = \frac { 1 } { 7 }$
$\Rightarrow \frac { \frac { 4 } { 3 } \pi r _ { 1 } ^ { 3 } } { \frac { 4 } { 3 } \pi r _ { 2 } ^ { 3 } } = \frac { 1 } { 7 } \Rightarrow \frac { 3 ^ { 3 } } { r _ { 2 } ^ { 3 } } = \frac { 1 } { 7 } \Rightarrow r _ { 2 } ^ { 3 } = 27 \times 7 = 189.$
Sum of volumes of two small spheres
$= \frac { 4 } { 3 } \pi \left( r _ { 1 } ^ { 3 } + r _ { 2 } ^ { 3 } \right) = \left[ \frac { 4 } { 3 } \pi \left( 3 ^ { 3 } + 189 \right) \right] \mathrm { cm } ^ { 3 }$
$= \left( \frac { 4 } { 3 } \pi \times 216 \right) \mathrm { cm } ^ { 3 }$
Let the radius of the new sphere be R.
Volume of new sphere = sum of volumes of two small spheres
$\Rightarrow \frac { 4 } { 3 } \pi R ^ { 3 } = \frac { 4 } { 3 } \pi \times 216 \Rightarrow R ^ { 3 } = 216$
$\Rightarrow R = \sqrt [ 3 ] { 216 } = 6 \mathrm { cm }$ $\Rightarrow$ diameter = 2R = 12cm.