When the force applied on PQ by long straight wire current carrying of $25 \mathrm{~A}$ rests on a table must balance the weight of small current carrying wire.
Then the magnetic field produced by long straight wire carrying current of $25 \mathrm{~A}$ rests on a table on small wire
$$
\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}}
$$
The magnetic force on small conductor is
$$
\begin{aligned}
\mathrm{F} &=\mathrm{BIl} \sin \theta \quad\left(\because \theta=90^{\circ}, \sin 90^{\circ}=1\right) \\
&=\mathrm{BIl}\left(\sin 90^{\circ}\right)=\mathrm{Bil}
\end{aligned}
$$
Consider the wire is balanced at hight $\mathrm{h}$ thus, the force applied on PQ with the weight of small current carrying wire.
$$
\begin{aligned}
&\mathrm{F}=\mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \quad\left(\therefore \mathrm{I}_1=\mathrm{I}_2=\mathrm{I}=25 \mathrm{~A}\right) \\
&\mathrm{h}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{mg}}=\frac{4 \pi \times 10^{-7} \times 25 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=51 \times 10^{-4}
\end{aligned}
$$
$\mathrm{h}=0.51 \mathrm{~cm}$