As per the question given,
The height of the cone $=28cm$
Surface area of the solid metallic sphere $=616c{{m}^{3}}$
As we know that,
Surface area of the sphere $=4\pi {{r}^{2}}$
Then, $4\pi {{r}^{2}}=616$
${{r}^{2}}=49$
$r=7$
Radius of the solid metallic sphere $=7cm$
Now assume r to be the radius of the cone
As we know that,
Volume of the cone $=1/3\pi {{r}^{2}}h$
$=1/3\pi {{r}^{2}}\left( 28 \right)$ ….. (i)
Volume of the sphere $=4/3\pi {{r}^{3}}$
$=4/3\pi {{7}^{3}}$ ………. (ii)
On equating equations (i) and (ii), we have
$1/3\pi {{r}^{2}}\left( 28 \right)=4/3\pi {{7}^{3}}$
Eliminating the common terms, we get
${{r}^{2}}\left( 28 \right)=4\times {{7}^{3}}$
${{r}^{2}}=49$
$r=7$
So, diameter of the cone $=7\times 2=14cm$
Therefore, the diameter of the base of the cone is $14cm$