The magnetic field due to wire $D$ at wire $C$ is
$B_{D } = \left(\frac{\mu_{0}}{4\pi}\right) \frac{2I}{r} $
$= \frac{10^{-7} \times 2\times30}{0.03} = 2\times10^{-4} T$
which is directed into the page.
The magnetic field due to wire $G$ at $C$ is
$B_G = \frac{10^{-7} \times 2 \times 20}{0.02} = 2 \times 10^{-4} \, T$
which is directed out of the page.
Therefore, the net magnetic field at the position of the wire $C$ is
$B = B_G - B_D = 2 \times 10^{-4} - 2\times 10^{-4} $ = zero.
The force on 25 cm of wire $C$ is
$F = BIl\, \sin \, \theta$ = zero.
