Step 1: Get the thermal diffusivity of the sphere material.
$\alpha = \dfrac{k}{\rho c} = \dfrac{10}{3000 \times 1000} = 3.333 \times 10^{-6}$ m$^2$/s.
This combines conduction, density and heat capacity into one property that sets how fast heat spreads inside the sphere.
Step 2: Find the Biot number to justify a uniform sphere temperature.
The characteristic length for a sphere is $L_c = r/3 = 0.005/3 = 1.667 \times 10^{-3}$ m.
$Bi = \dfrac{hL_c}{k} = \dfrac{500 \times 1.667 \times 10^{-3}}{10} = 0.0833$, small enough (under 0.1) to treat the sphere as isothermal at each instant.
Step 3: Write cooling in terms of Biot and Fourier numbers.
For lumped cooling, $\dfrac{T-T_\infty}{T_i-T_\infty} = e^{-Bi \cdot Fo}$, where the Fourier number is $Fo = \dfrac{\alpha t}{L_c^2}$.
With $T_i = 400^{\circ}$C, $T_\infty = 20^{\circ}$C, $T = 50^{\circ}$C, the ratio is $\dfrac{50-20}{400-20} = \dfrac{30}{380} = 0.0789$.
Step 4: Solve for the Fourier number, then the time.
$\ln(1/0.0789) = 2.539 = Bi \cdot Fo = 0.0833 \times Fo$, so $Fo = 30.47$.
$t = \dfrac{Fo \cdot L_c^2}{\alpha} = \dfrac{30.47 \times (1.667\times10^{-3})^2}{3.333\times10^{-6}} = 25.40$ s.
Final Answer:
Both the direct time constant route and this Biot Fourier route land on the same cooling time.
\[ \boxed{t \approx 25.39 \text{ s}} \]