Account 1 :
[tex]Interst= \dfrac{p\times r\times t}{100}\\\\2 = \dfrac{p\times 3.2\times 3}{100}\\\\p = \$20.83[/tex]
Account 2 :
[tex]Interst= \dfrac{p\times r\times t}{100}\\\\38.50 = \dfrac{p\times 2.2\times 30}{100}\\\\p = \$58.33[/tex]
Interest of first month :
1 ) [tex]I=\dfrac{20.83 \times 3.2}{100}=\$0.67[/tex]
2) [tex]I=\dfrac{58.33 \times 2.2}{100}=\$1.28[/tex]
Account 2 has most interest in 1 st month.
Hence, this is the required solution.