Answer
4.3 A
Work Step by Step
Magnetic force on a current-carrying wire is
$F=ILB\sin\theta$
$\implies 0.030\,N=(2.7\,A)LB\sin\theta$
Or $LB\sin\theta =\frac{0.030\,N}{2.7\,A}=0.011\,N/A$
The new force
$F'=0.047\,N=I'LB\sin\theta$ where $I'$ is the new current.
$L$, $B$ and $\theta$ are same
$\implies I'=\frac{F'}{LB\sin\theta}=\frac{0.047\,N}{0.011\frac{N}{A}}=4.3\,A$
