Each year, the initial deposit gets multiplied by a factor of:
[tex](1+\frac{5}{100})[/tex]Let L be the initial deposit. 6 years later, the balance of the account will be equal to:
[tex]L\cdot(1+\frac{5}{100})^6[/tex]On the other hand, the current balance is $300. Therefore:
[tex]L\cdot(1+\frac{5}{100})^6=300[/tex]Solve for L:
[tex]\begin{gathered} L=\frac{300}{(1+\frac{5}{100})^6} \\ =\frac{300}{1.05^6} \\ =223.8646\ldots \\ \cong223.86 \end{gathered}[/tex]Therefore, the initial amount of money in the account 6 years ago, was:
[tex]223.86[/tex]