A steady current I enters a circuit at node P and leaves at node Q...

The circuit consists of two parallel branches between nodes $P$ and $Q$.

The resistance of the upper branch is $R_{upper} = R_1$.

The resistance of the lower branch is $R_{lower} = R_2 + R_3$, as the two resistors are in series.

Since the branches are in parallel, the potential difference $V$ across both branches is the same.

The current through the lower branch is $I_2 = \dfrac{V}{R_2 + R_3}$.

The total current $I$ is given by $I = \dfrac{V}{R_{eq}}$, where $R_{eq}$ is the equivalent resistance of the circuit.

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2 + R_3} = \dfrac{R_1 + R_2 + R_3}{R_1(R_2 + R_3)}$

$\Rightarrow R_{eq} = \dfrac{R_1(R_2 + R_3)}{R_1 + R_2 + R_3}$

Therefore, $I = \dfrac{V(R_1 + R_2 + R_3)}{R_1(R_2 + R_3)}$.

Taking the ratio of $I_2$ to $I$:

$\dfrac{I_2}{I} = \dfrac{\dfrac{V}{R_2 + R_3}}{\dfrac{V(R_1 + R_2 + R_3)}{R_1(R_2 + R_3)}} = \dfrac{R_1}{R_1 + R_2 + R_3}$

Answer: $\dfrac{R_1}{R_1 + R_2 + R_3}$