Given that, at a distance $r_1 = 5 \text{ cm} = 0.05 \text{ m}$ from the center of an electric dipole on its axis, the electric field intensity is $E_1 = 2.5 \times 10^4 \text{ NC}^{-1}$. At a distance $r_2 = 10 \text{ cm} = 0.10 \text{ m}$, the electric field intensity is $E_2 = 2 \times 10^3 \text{ NC}^{-1}$. The formula for the electric field intensity $E$ at a point on the axis of an electric dipole is given by: $$E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2pr}{(r^2 - l^2)^2}$$ where $p = q \cdot 2l$ is the dipole moment and $2l$ is the length of the dipole. According to the problem: $$\begin{aligned} E_1 &= \frac{1}{4 \pi \epsilon_0} \cdot \frac{2p(0.05)}{(0.05^2 - l^2)^2} \quad \dots (i) \\ E_2 &= \frac{1}{4 \pi \epsilon_0} \cdot \frac{2p(0.10)}{(0.10^2 - l^2)^2} \quad \dots (ii) \end{aligned}$$ Dividing equation $(i)$ by equation $(ii)$, we get: $$\begin{aligned} \frac{E_1}{E_2} &= \frac{0.05}{(0.05^2 - l^2)^2} \cdot \frac{(0.10^2 - l^2)^2}{0.10} \\ \Rightarrow \frac{2.5 \times 10^4}{2 \times 10^3} &= \frac{1}{2} \cdot \left( \frac{0.01 - l^2}{0.0025 - l^2} \right)^2 \\ \Rightarrow 12.5 &= \frac{1}{2} \cdot \left( \frac{0.01 - l^2}{0.0025 - l^2} \right)^2 \\ \Rightarrow 25 &= \left( \frac{0.01 - l^2}{0.0025 - l^2} \right)^2 \\ \Rightarrow 5 &= \frac{0.01 - l^2}{0.0025 - l^2} \\ \Rightarrow 0.0125 - 5l^2 &= 0.01 - l^2 \\ \Rightarrow 4l^2 &= 0.0025 \\ \Rightarrow l^2 &= 0.000625 \\ \Rightarrow l &= 0.025 \text{ m} \end{aligned}$$ Therefore, the length of the electric dipole is: $$2l = 2 \times 0.025 \text{ m} = 0.05 \text{ m}$$ The length of the electric dipole is $0.05 \text{ m}$.