As given that, for ward biased resistance $=25 \Omega$
Reverse biased resistance $=\infty, \mathrm{V}=5$ volt
As the diode in branch $\mathrm{CD}$ is in reverse biased which having resistance infinite,
So $\left(I_3=0\right)$
Diode $\mathrm{D}_1$ and $\mathrm{D}_3$ are in forward bias $\mathrm{D}_2$ is in reverse bias, So, Resistance in branch $\mathrm{AB}=25+125=150 \Omega$ say $\mathrm{R}_1$ Resistance in branch $\mathrm{EF}=25+125=150 \Omega$ say $\mathrm{R}_2$ $\mathrm{AB}$ is parallel to $\mathrm{EF}$.
So, Net resistance $\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}$
$$
\Rightarrow \mathrm{R}^{\prime}=75 \Omega
$$
Total resistance $\mathrm{R}=\mathrm{R}^{\prime}+25=75+25=100 \Omega$
So, $\mathrm{V}=\mathrm{IR}$, current $\mathrm{I}_1=\frac{\mathrm{V}}{\mathrm{R}}=\frac{5}{100}=0.05 \mathrm{Amp}$
As per figure $I_1=I_4+I_2+I_3 \quad\left(\because I_3=0\right)$
So, $I_1=I_4+I_2$
Here resistance $\mathrm{R}_1$ and $\mathrm{R}_2$ are same.
i.e., $I_4=I_2$
So, $\mathrm{I}_1=2 \mathrm{I}_2$
$$
\Rightarrow \quad \mathrm{I}_2=\frac{\mathrm{I}_1}{2}=\frac{0.05}{2}=0.025 \mathrm{Amp}
$$
and $I_2=0.025 \mathrm{Amp}, I_4=0.025 \mathrm{Amp}\left(\therefore \mathrm{I}_4=\mathrm{I}_2\right)$
So, $\mathrm{I}_1=0.05 \mathrm{Amp} \mathrm{I}_2=0.025 \mathrm{Amp}, \mathrm{I}_3=0$
and $\mathrm{I}_4=0.025 \mathrm{Amp}$