Step 1: Use magnetic field from a finite wire.
For each side of the triangle:
$B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$.
Step 2: Geometry of equilateral triangle.
Distance from centroid to each side = $\frac{\sqrt{3}}{6}a$.
Angles are $60^\circ$ each: $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
Step 3: Field of one side.
$B_{1} = \frac{\mu_0 I}{4\pi (\frac{\sqrt{3}}{6}a)} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right)$
$B_{1} = \frac{\mu_0 I}{4\pi a} \cdot 3$.
Step 4: Total field (three sides).
$B_{\text{total}} = 3\cdot \frac{\mu_0 I}{4\pi a} \cdot 3 = \frac{3\mu_0 I}{2\pi a}$.
Conclusion. The magnetic field at the centroid is $\dfrac{3\mu_{0}I}{2\pi a}$.

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)
