The visible surface of the Sun is at about 5800 K…

Step 1. The energy. $\bar{K} = \tfrac32 \times 1.38 \times 10^{-23}\ \text{J/K} \times 5800\ \text{K} = 1.20 \times 10^{-19}$ J.

Step 2. In eV. $\dfrac{1.20 \times 10^{-19}\ \text{J}}{1.6 \times 10^{-19}\ \text{J/eV}} = 0.75$ eV.

Step 3. The temperature for 1.0 eV. We want $T$, so $T = \dfrac{2\bar{K}}{3k_B} = \dfrac{2 \times 1.6 \times 10^{-19}\ \text{J}}{3 \times 1.38 \times 10^{-23}\ \text{J/K}} = 7730$ K.