The root mean square of the current is given by
![I_(rms) = (I_0)/( √(2) )](https://img.qammunity.org/2019/formulas/physics/middle-school/pd8ajk7wd4l1bwrk9rv3hnh8zzoniin43g.png)
where
![I_0](https://img.qammunity.org/2019/formulas/physics/middle-school/ihg2sfqzr5ls0chkxxdc812hspdsmodgjz.png)
is the maximum value of the current.
In our problem, the maximum current allowed without breaking the filament is equal to
![I_0=1.50 A](https://img.qammunity.org/2019/formulas/physics/college/v56vmkh5r596q2z6fh3vrn79g778ho0bos.png)
. Therefore, the largest root-mean-square current allowed without breaking the wire is
![I_(rms)= (1.50 A)/( √(2) )=1.06 A](https://img.qammunity.org/2019/formulas/physics/college/atwloy6ia76bdef8848nag6eu99icssjds.png)