1. Given data

   In triangle $ABC$, $a=6,\, b=10$ and area $\Delta = 15\sqrt{3}.$

   Also, $\angle C = \angle ACB$ is obtuse.

   We need to find $r^2$, where $r$ is the inradius.

2. Use area formula with two sides and included angle

   Since sides adjacent to angle $C$ are $a$ and $b$, the area is $\Delta = \frac12 ab\sin C.$

   Substitute the values: $15\sqrt{3} = \frac12 \cdot 6 \cdot 10 \cdot \sin C$ $15\sqrt{3} = 30\sin C$ $\sin C = \frac{\sqrt{3}}{2}.$

3. Determine angle $C$

   Since $\sin C = \frac{\sqrt{3}}{2}$, possible angles are $C = 60^\circ \quad \text{or} \quad 120^\circ.$

   Given that $\angle C$ is obtuse, we take $C = 120^\circ.$

4. Find side $c$ using the cosine rule

   $c^2 = a^2 + b^2 - 2ab\cos C.$

   Since $\cos 120^\circ = -\frac12,$ we get $c^2 = 6^2 + 10^2 - 2\cdot 6 \cdot 10 \left(-\frac12\right)$ $c^2 = 36 + 100 + 60 = 196$ $c=14.$

5. Find semiperimeter

   $s = \frac{a+b+c}{2} = \frac{6+10+14}{2} = 15.$

6. Use the area formula $\Delta = rs$

   $15\sqrt{3} = r \cdot 15$ $r = \sqrt{3}.$

7. Compute $r^2$

   $r^2 = (\sqrt{3})^2 = 3.$

Therefore, the required value is $\boxed{3}.$