Change in volume of $1 \,kg$ of water due to phase change is

$\Delta V=V_{\text {steam }}-V_{\text {water }}$

$=\frac{m_{\text {seam }}-m_{\text {water }}}{\rho_{\text {steam }}}-\rho_{\text {water }}$

$=\frac{1}{(1 / 18)}-\frac{1}{1000}$

$=1800-0.001=1799 \,m ^3$

Work done against atmospheric pressure during phase change is

$\Delta W =p \Delta V$

$=101 \times 10^5 \times 1799$

$=181699 \,J$

Heat absorbed during phase change is

$\Delta Q =m L$

$=1 \times 22.6 \times 10^5$

$=2260000 \,J$

So, change in internal energy is

$\Delta U =\Delta Q-\Delta W$

$=2260000-181699$

$=2078301 J \,kg$

$=208 \times 10^5 \,J kg ^{-1}$