Answer: Option B
Volume of new sphere :
$$\eqalign{
& = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr
& = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr
& = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr
& = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the new sphere be R
Then,
$$\eqalign{
& \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr
& \Rightarrow R = 12\,cm \cr} $$
∴ Diameter :
$$\eqalign{
& = 2R \cr
& = 2 \times 12 \cr
& = 24\,cm \cr} $$