4. Given that the radius of the circumcircle of a triangle is $4 \mathrm{~cm}$, one of its internal angles is $60^{\circ}$, and the difference between the two sides enclosing this angle is $4 \mathrm{~cm}$, then, the area of this triangle is $\mathrm{cm}$.

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4. Given that the radius of the circumcircle of a triangle is $4 \mathrm{~cm}$, one of its internal angles is $60^{\circ}$, and the difference between the two sides enclosing this angle is $4 \mathrm{~cm}$, then, the area of this triangle is $\mathrm{cm}$.
4. $8 \sqrt{3}$

Solution Without loss of generality, let $\triangle A B C$ have $\angle A=60^{\circ}, b-c=4, R=4$. By the cosine and sine theorems, we get
$$
a^{2}=b^{2}+c^{2}-2 b c \cos 60^{\circ}=(b-c)^{2}+b c=16+b c,
$$

and $a=2 R \sin \angle A=4 \sqrt{3}$.
Then $b c=a^{2}-16=32$.
$$
S=\frac{1}{2} b c \sin A=8 \sqrt{3} \text {. }
$$