The density of a material in \(SI\, units\) is \(128\, kg\, m^{-3}\). In certain units in which the unit of length is \(25\, cm\) and the unit of mass \(50\, g\), the numerical value of density of the material is

Explanation:
To convert the numerical value of density from one system of units to another, the principle of dimensional analysis is applied:
\( n_1 u_1 = n_2 u_2 \)
where \( n \) represents the numerical value and \( u \) represents the unit configuration.

1. Identify the Dimensions of Density:
The dimensional formula for density (\( \rho \)) is expressed as:
\( [\text{Density}] = [M^1 L^{-3} T^0] \)

2. Set up the Conversion Equation:
Using the powers of the fundamental quantities from the dimensional formula, the relation between the two systems is:
\( n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^1 \left[ \frac{L_1}{L_2} \right]^{-3} \)

3. Substitute the Corresponding Values:
- SI System (System 1):
- \( n_1 = 128 \)
- \( M_1 = 1\text{ kg} = 1000\text{ g} \)
- \( L_1 = 1\text{ m} = 100\text{ cm} \)

- New System (System 2):
- \( M_2 = 50\text{ g} \)
- \( L_2 = 25\text{ cm} \)

Plugging these variables into the conversion formula yields:
\( n_2 = 128 \times \left[ \frac{1000\text{ g}}{50\text{ g}} \right]^1 \times \left[ \frac{100\text{ cm}}{25\text{ cm} } \right]^{-3} \)

4. Simplify the Terms:
- Ratio for Mass: \( \frac{1000}{50} = 20 \)
- Ratio for Length: \( \frac{100}{25} = 4 \)

Substituting these simplified values back into the equation:
\( n_2 = 128 \times (20) \times (4)^{-3} \)
\( n_2 = 128 \times 20 \times \frac{1}{64} \)

Since \( \frac{128}{64} = 2 \):
\( n_2 = 2 \times 20 = 40 \)

Hence, Option A is the correct choice.