Assuming the Sun radiates as an ideal blackbody at 6000 K, what intensity (dP/dA/dλ) does Planck’s law predict at λ = 300 nm, and what is the approximate emitted power per unit area in the UVB range 285–315 nm using dP/dA/dλ × Δλ?

What you are estimating

You want the Sun’s emitted blackbody power in a narrow wavelength band (UVB, 285–315 nm). Since the band is only $30\,\text{nm}$ wide, you are allowed to approximate the integral by taking the spectral value at the center ($300\,\text{nm}$) and multiplying by $\Delta\lambda$.

Planck law in the form that matches $dP/(dA\, d\lambda)$

For an ideal blackbody, the hemispherical spectral exitance (also called spectral emissive power) is $$ M_\lambda=\frac{dP}{dA\, d\lambda}=\frac{2\pi h c^2}{\lambda^5}\,\frac{1}{\exp\left(\frac{hc}{\lambda kT}\right)-1}. $$ Here $h$ is Planck’s constant, $c$ is the speed of light, and $k$ is Boltzmann’s constant.

Evaluate the exponential term at $\lambda=300\,\text{nm}$, $T=6000\,\text{K}$

Compute $$ x=\frac{hc}{\lambda kT}. $$ Using $hc\approx 1.986\times 10^{-25}\,\text{J·m}$ and $\lambda kT=(3.00\times 10^{-7})(1.380649\times 10^{-23})(6000)$, $$ x\approx 8.0, \qquad e^x-1\approx 3.0\times 10^3. $$

Compute $M_\lambda$ at 300 nm and convert to per-nanometer

First compute the prefactor: $$ \frac{2\pi h c^2}{\lambda^5} \approx \frac{3.74\times 10^{-16}}{(3.00\times 10^{-7})^5} \approx 1.54\times 10^{17}\,\text{W m}^{-3}. $$ Then divide by $e^x-1$: $$ M_\lambda(300\,\text{nm})\approx \frac{1.54\times 10^{17}}{3.0\times 10^{3}} \approx 5.2\times 10^{13}\,\text{W m}^{-3}. $$ Since $1\,\text{nm}=10^{-9}\,\text{m}$, $$ M_\lambda \approx 5.2\times 10^{4}\,\text{W m}^{-2}\,\text{nm}^{-1}. $$

Approximate the UVB band power using $M_\lambda\,\Delta\lambda$

With $\Delta\lambda=315-285=30\,\text{nm}$, $$ \left(\frac{P}{A}\right)_{\text{UVB}} \approx M_\lambda(300\,\text{nm})\,\Delta\lambda \approx (5.2\times 10^{4})(30) \approx 1.6\times 10^{6}\,\text{W m}^{-2}. $$

Quick reasonableness check

The total blackbody exitance at $6000\,\text{K}$ is $\sigma T^4\approx 7.3\times 10^7\,\text{W m}^{-2}$, so this UVB estimate is a few percent of the total, which is plausible for a $6000\,\text{K}$ spectrum.