A spherical ball of radius 3cm …

Radius of the ball = {tex}\frac { 3 } { 2 } \mathrm { cm } = 1.5 \mathrm { cm }{/tex}
Volume of the bigger ball =  {tex}\frac{4}{3}\pi {(1.5)^3}{\text{c}}{{\text{m}}^3} {/tex}...... (i)
Radii of two smaller balls are {tex}\frac { 1 } { 2 } \mathrm { cm } \text { and } \frac { 1.5 } { 2 } \mathrm { cm }{/tex}
{tex}\therefore {/tex} total volume of the two smaller balls
{tex}= \left[ \frac { 4 } { 3 } \pi ( 0.5 ) ^ { 3 } + \frac { 4 } { 3 } \pi ( 0.75 ) ^ { 3 } \right] \mathrm { cm } ^ { 3 }{/tex}
{tex}= \frac{4}{3}\pi \left[ {{{(0.5)}^3} + {{(0.75)}^3}} \right]{\text{c}}{{\text{m}}^3} {/tex}.......(ii)
Let the radius of the third ball be r cm.
{tex}\therefore {/tex} Volume of the third ball = {tex}\frac{4}{3}\pi {r^3}c{m^3}\, {/tex}......... (iii)

Since, the big spherical ball is melted to produce the three small spherical balls; the volume of the big spherical ball is same as the sum of the volumes of the three small spherical balls. 
Therefore, we have

From (i), (ii) and (iii), we have
{tex}\frac { 4 } { 3 } \pi ( 1.5 ) ^ { 3 } = \frac { 4 } { 3 } \pi \left[ ( 0.5 ) ^ { 3 } + ( 0.75 ) ^ { 3 } \right] + \frac { 4 } { 3 } \pi r ^ { 3 } \Rightarrow ( 1.5 ) ^ { 3 } = ( 0.5 ) ^ { 3 } + ( 0.75 ) ^ { 3 } + r ^ { 3 }{/tex}
{tex}\Rightarrow r ^ { 3 } = ( 1.5 ) ^ { 3 } - ( 0.5 ) ^ { 3 } - ( 0.75 ) ^ { 3 } = r ^ { 3 } = 3.375 - 0.125 - 0.421875{/tex}
{tex}\Rightarrow r ^ { 3 } = 2.828125 \Rightarrow r = 1.41{/tex}
{tex}\therefore {/tex} radius of the third ball = 1.41 cm