The root mean square of the current is given by
[tex]I_{rms} = \frac{I_0}{ \sqrt{2} } [/tex]
where [tex]I_0[/tex] is the maximum value of the current.
In our problem, the maximum current allowed without breaking the filament is equal to [tex]I_0=1.50 A[/tex]. Therefore, the largest root-mean-square current allowed without breaking the wire is
[tex]I_{rms}= \frac{1.50 A}{ \sqrt{2} }=1.06 A [/tex]
