When resistors are connected in parallel potential difference across them is same. In the given circuit the resistor's $R_{2}$ and $R_{3}$ are connected in parallel hence potential difference $(V)$ across them is same. In order that they undergo same energy loss, $H=\frac{V^{2}}{R} t$,
$R_{2}$ must be equal to $R_{3}$.
i.e., $R_{2}=R_{3}$
Now resistor $R_{1}$ is in series with $R_{2}$, hence energy through them is
$H=i^{2} R_{1} t=i_{1}^{2} R_{2} t$
where $i_{1}$ is current across $R_{2}$.
Since $R_{2}=R_{3}$, therefore current through them is $\frac{i}{2}=i_{1}$
$\therefore i^{2} R_{1} t=\frac{i^{2}}{4} R_{2} t$
$\Rightarrow R_{1}=\frac{R_{2}}{4}$