As we know that resistance of the resistor will remain same while connecting across different batteries
So here we will say by ohm's law
[tex]R = \frac{V}{i}[/tex]
here we know that
[tex]i = 0.60 A[/tex]
[tex]V = 373 Volts[/tex]
now we have
[tex]R = \frac{373}{0.60} = 621.67 ohm[/tex]
now we have
another voltage as V = 139 volts
now we have current
[tex]i = \frac{V}{R} = \frac{139}{621.67}[/tex]
[tex]i = 0.22 A[/tex]
so final current is 0.22 A