b^{-10} \cdot \left(\frac{5b^4}{3b^{12}}\right)^3 = 125b^{-2} = -0,8

1. Simplify the expression: $$b^{-10} \cdot \frac{(5b^4)^3}{(3b^{12})^3} = b^{-10} \cdot \frac{125b^{12}}{27b^{36}} = \frac{125b^2}{27b^{26}} = \frac{125}{27b^{24}}$$.
2. Equate to the given value: $$\frac{125}{27b^{24}} = -0.8$$.
3. Solve for $$b^{24}$$: $$125 = -0.8 \cdot 27b^{24} \implies b^{24} = \frac{125}{-0.8 \cdot 27} = \frac{125}{-21.6}$$. This leads to no real solution for $$b$$.