Answer
$I_2=7.848\times10^{-5}A$
Work Step by Step
$I_1=I_2+I_3+I_4$
$0=I_1R_1-E_1+I_4R_4$
$21V=(13000\Omega)I_1+(18000\Omega)I_4$
$0=-I_4R_4-E_3+I_3R_3$
$12V=(12000\Omega)I_3-(18000\Omega)I_4$
$16V=(12000\Omega)I_3-(10000\Omega)I_2$
$17V=(13000\Omega)I_1+(10000\Omega)I_2$
$I_4=\frac{21V-(13000\Omega)I_1}{18000\Omega}$
$I_3=\frac{12V+(18000\Omega)I_4}{12000\Omega}=
\frac{33V-(13000\Omega)I_1}{12000\Omega}$
$I_2=\frac{17V-(13000\Omega)I_1}{10000\Omega}$
$I_1=\Big(\frac{17V-(13000\Omega)I_1}{10000\Omega}\Big)+\Big(\frac{33V-(13000\Omega)I_1}{12000\Omega}\Big)+\Big(\frac{21V-(13000\Omega)I_1}{18000\Omega}\Big)=0.00137A$
$17=(13000\Omega\times1.368\times10^{-3})+(10000\Omega)I_2$
$I_2=7.848\times10^{-5}A$