Radiation

Surface area of centrifugal rod =\pi d h+\frac{2 \pi d^{2}}{4}
Or A_{1}=\pi\left(d h+\frac{d^{2}}{2}\right)
Surface area of cube A_{2}=6 \times \mathrm{L}^{2}

For surface 1 (cylinder), \mathrm{F}_{11}=0
So \mathrm{F}_{12}=1 \quad \text {...(i) }

For surface 2(cube)
\begin{aligned} \mathrm{F}_{21}+\mathrm{F}_{22} & =1 \\ \mathrm{F}_{22} & =1-\mathrm{F}_{21} \\ \mathrm{F}_{21} & =1-\mathrm{F}_{22}\;\;...(ii) \end{aligned}
By reciprocity theorem for surface 1 and 2
\begin{aligned} A_{1} F_{12} & =A_{2} F_{21} \\ F_{21} & =\frac{A_{1}}{A_{2}} \end{aligned}
\left[F_{12}=1\right. from equation (i)]
So, By equation (ii)
F_{22}=F_{s S}=1-\left(\frac{\pi \mathrm{dh}+\frac{\pi d^{2}}{2}}{6 L^{2}}\right)

From the figure we can say that at 6000 \mathring{A} wavelength total hemispherical emissivity is maximum
\lambda _m=6000 \mathring{A} =0.6\mu m
Wein's law
\lambda _m T=C
\lambda _m T=2898 \mu mK
T=\frac{2898}{0.6}=4830K
Note: In this question, we need to find temperature at which total hemispherical emissivity will be highest.
But to calculate total hemispherical emissivity we require additional data like function chart and emissivity value at particular wavelength.
With the use of Wein's law we can find out temperature at which spectral emissivity is highest but in question it is asked temperature for total hemispherical emissivity.

Equilibrium Temperature =T_S=?
In this it is mentioned that entire flux is absorbed by the plate it means for solar flux absorptivity is 1. Kirchoff's law \alpha =\varepsilon =0.21
Surface of cast iron and surrounding fluid temperature difference is small due to this we can use Kirchoff's law
At equilibrium condition
Energy absorbed = Energy leaving
\begin{aligned} \alpha GA&=\varepsilon A\sigma (T_S^4-T_\infty ^4)\\ 1 \times 600 &=0.21 \times 5.669 \times 10^{-8}(T_S^4-298^4)\\ T_S&=491.34K\\ T_S&=218.34^{\circ}C\\ T_S&=218^{\circ}C \end{aligned}

\begin{aligned} r_{1} &=10 \mathrm{~mm} \\ A_{1} &=4 \pi r_{1}^{2} \\ A_{2} &=6 \times\left(30^{2}\right) \\ F_{12} &=1 \\ A_{1} F_{12} &=A_{2} F_{21} \quad \Rightarrow \quad F_{21}=\frac{A_{1}}{A_{2}}=\frac{4 \pi \times(10)^{2}}{6 \times(30)^{2}}=0.2327 \\ F_{22} &=1-F_{21}=0.7672 \simeq 0.77 \end{aligned}

From Wien's Displacement law,
\begin{aligned} \lambda_{m} T &=\text { constant } \Rightarrow \lambda_{m 1} T_{1}=\lambda_{m 2} T_{2} \\ \lambda_{m 1} \times 3000 &=1.2 \lambda_{m 1} \times T_{2} \\ T_{2} &=\left(\frac{3000}{1.2}\right) K=2500 K \end{aligned}