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 $14cm$