detailed solution

Correct option is Ğ

$$\begin{gathered} {\mathrm{w}}_0=4.41\times {10}^{-19}\mathrm{J} \\ \mathrm{\lambda }=300\mathrm{nm}, \mathrm{K}.\mathrm{E}=?\mathrm{h}=6.63\times {10}^{-34}\mathrm{Js}, \\ {\mathrm{w}}_0=\mathrm{h\upsilon }-\mathrm{K}.\mathrm{E}. \\ 4.41\times {10}^{-19}\mathrm{J}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{300\times {10}^{-9}}-\mathrm{K}.\mathrm{E} \\ 4.41\times {10}^{-19}\mathrm{J}=\frac{19.89\times {10}^{-26}}{300\times {10}^{-9}}-\mathrm{K}.\mathrm{E} \\ 4.41\times {10}^{-19}\mathrm{J}=6.63\times {10}^{19}-\mathrm{K}.\mathrm{E} \\ \mathrm{K}.\mathrm{E}=2.22\times {10}^{-19}\mathrm{J} \\ =222\times {10}^{-21}\mathrm{J} \end{gathered}$$

detailed solution

Correct answer is 222

$$\begin{gathered} {\mathrm{w}}_0=4.41\times {10}^{-19}\mathrm{J} \\ \mathrm{\lambda }=300\mathrm{nm}, \mathrm{K}.\mathrm{E}=?\mathrm{h}=6.63\times {10}^{-34}\mathrm{Js}, \\ {\mathrm{w}}_0=\mathrm{h\upsilon }-\mathrm{K}.\mathrm{E}. \\ 4.41\times {10}^{-19}\mathrm{J}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{300\times {10}^{-9}}-\mathrm{K}.\mathrm{E} \\ 4.41\times {10}^{-19}\mathrm{J}=\frac{19.89\times {10}^{-26}}{300\times {10}^{-9}}-\mathrm{K}.\mathrm{E} \\ 4.41\times {10}^{-19}\mathrm{J}=6.63\times {10}^{19}-\mathrm{K}.\mathrm{E} \\ \mathrm{K}.\mathrm{E}=2.22\times {10}^{-19}\mathrm{J} \\ =222\times {10}^{-21}\mathrm{J} \end{gathered}$$