If each diode has a forward bias resistance of 25Ω in the below...

As the diode ${D}_{2}$ is connected in reverse order, there is no current through it.

The effective resistances of the circuit can, then, be calculated as follows:

$\begin{cases}{R}_{eff} & = & 25 + \dfrac{{\left(25 + 125\right)}^{2}}{\left(25 + 125\right) + \left(25 + 125\right)}Ω \\ & = & 100Ω\end{cases}$

Each current through the diodes can be calculated as follows:

$\begin{cases}{I}_{1} & = & \left(\dfrac{5V}{100Ω}\right) \\ & = & 0 . 05A\end{cases}$

As the diode ${D}_{2}$ is non-conducting, the current ${I}_{1}$ will be divided equally among the other two diodes ${D}_{1}$ and ${D}_{3}$.

Hence, the currents in the diodes can be written as

$\begin{cases}{I}_{2} & = & \dfrac{0 . 05A}{2} \\ & = & 0 . 025A \\ {I}_{4} & = & 0 . 025A\end{cases}$

Thus,

$\begin{cases}\dfrac{{I}_{1}}{{I}_{2}} & = & \dfrac{0 . 05A}{0 . 025A} \\ & = & 2\end{cases}$