Ionization energy for hydrogen atom in ergs, joules and ev respectively is:

$$\begin{gathered} ∆\mathrm{E} =\frac{-13.6 \times {\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{ev} (\mathrm{n}=1 ,\mathrm{Z}=1) \\ =-13.6\mathrm{ev} \\ 1\mathrm{ev} =1.6 \times {10}^{-19}\mathrm{jouls} \\ \mathrm{so} ∆\mathrm{E}=13.6\times 1.6\times {10}^{-19} \\ =218\times {10}^{-20 }\mathrm{jouls} \\ 1\mathrm{joule} ={10}^7\mathrm{ergs} \\ \mathrm{so}∆\mathrm{E}=218\times {10}^{-20}\times {10}^7 \\ ∆\mathrm{E} =218\times {10}^{-12}\mathrm{ergs} \end{gathered}$$