In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, respectively. If $\mu_{0}$ and $\epsilon_{0}$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{\mu_{0}/\epsilon_{0}},$ is:

Step 1: Define the base units of the new system.

The units of mass, length, time, and current in the new system are given as:

$$M_0 = 5\,\text{kg}$$ $$L_0 = 5\,\text{m}$$ $$T_0 = 5\,\text{s}$$ $$I_0 = 5\,\text{A}$$

Step 2: Find the dimensions of permeability $\mu_0$.

Using the magnetic force per unit length between two parallel current-carrying conductors:

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$

This yields the dimensional formula:

$$[\mu_0] = [M L T^{-2} I^{-2}]$$

Step 3: Find the dimensions of permittivity $\varepsilon_0$.

Using Coulomb's Law for electrostatic force:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

Since charge $q = I \cdot T$, the dimensional formula is:

$$[\varepsilon_0] = [M^{-1} L^{-3} T^{4} I^{2}]$$

Step 4: Determine the dimensions of the quantity $\sqrt{\mu_0/\varepsilon_0}$.

Divide the dimensions of $\mu_0$ by $\varepsilon_0$:

$$\left[\frac{\mu_0}{\varepsilon_0}\right] = [M^{2} L^{4} T^{-6} I^{-4}]$$

Taking the square root gives the dimensions of the quantity:

$$\left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right] = [M L^{2} T^{-3} I^{-2}]$$

Step 5: Calculate the value of the new unit in terms of SI units.

Substitute the values of the new base units into the dimensional formula:

$$\text{New Unit} = (5)^{1} \cdot (5)^{2} \cdot (5)^{-3} \cdot (5)^{-2} \text{ SI units}$$ $$\text{New Unit} = 5^{1+2-3-2} \text{ SI units}$$ $$\text{New Unit} = 5^{-2} \text{ SI units}$$ $$\text{New Unit} = \frac{1}{25} \text{ SI units}$$

Step 6: Convert one SI unit into the new system of units.

Using the relation $n_1 u_1 = n_2 u_2$:

$$1 \text{ SI unit} = N \times (\text{New Unit})$$ $$1 = N \times \frac{1}{25}$$ $$N = 25$$

Answer: 25