1. Given density in SI units

   $\rho = 128\ \text{kg m}^{-3}$

2. New fundamental units

   • Unit of length: $L' = 25\ \text{cm} = 0.25\ \text{m}$
   • Unit of mass: $M' = 50\ \text{g} = 0.05\ \text{kg}$

3. Unit of density in the new system

   Density has dimensions: $[\rho] = M L^{-3}$

   So, one new unit of density is $\rho' = \frac{M'}{(L')^3}$

   Substitute the values: $\rho' = \frac{0.05}{(0.25)^3}$

   $(0.25)^3 = \frac{1}{64} = 0.015625$

   Hence, $\rho' = \frac{0.05}{0.015625} = 3.2\ \text{kg m}^{-3}$

4. Find the numerical value in the new system

   If the numerical value is $n$, then $128 = n \times 3.2$

   Therefore, $n = \frac{128}{3.2} = 40$

5. Option check

   • A: $40$ ✅
   • B: $640$ ❌
   • C: $16$ ❌
   • D: $410$ ❌

Therefore, the numerical value of density in the new units is:

$\boxed{40}$