$$6 \frac{1}{2} t^2 h^3 d \cdot (\frac{1}{2}th)^4 = \frac{13}{2} t^2 h^3 d \cdot (\frac{1}{2})^4 \cdot t^4 \cdot h^4 = \frac{13}{2} t^2 h^3 d \cdot \frac{1}{16} t^4 h^4 = \frac{13}{32} t^{2+4} h^{3+4} d = \frac{13}{32} t^6 h^7 d$$
Ответ: $$\frac{13}{32} t^6 h^7 d$$