Answer: Option A
Let the radius of the third ball be R cm
Then,
$$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$
$$\eqalign{
& \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr
& \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr
& \Rightarrow R = \frac{5}{4} \cr} $$
∴ Diameter of the third ball :
$$ = 2R = \frac{5}{2}cm = 2.5\,cm$$