Let $A$ be John's total weekly allowance.
1. Arcade Expense

John spent $\(\frac{3}{5}\)$ of his allowance at the arcade.

$$
\(\text { Arcade Expense }=\frac{3}{5} A\)
$$

2. Remaining Allowance after Arcade

The remaining fraction of his allowance is:

$$
\(\text { Remaining }_1=1-\frac{3}{5}=\frac{5}{5}-\frac{3}{5}=\frac{2}{5} A\)
$$

3. Toy Store Expense

The next day, he spent $\(\frac{1}{3}\)$ of his remaining allowance at the toy store.

$$
\(\text { Toy Store Expense }=\frac{1}{3} \times \text { Remaining }_1=\frac{1}{3} \times\left(\frac{2}{5} A\right)=\frac{2}{15} A\)
$$

4. Remaining Allowance after Toy Store

The money remaining after the toy store expense is the rest of Remaining $\({ }_1\)$. Since he spent $\(\frac{1}{3}\)$ of Remaining $\(_1\)$, he has $\(\frac{2}{3}\)$ of Remaining $\({ }_1\)$ left.

$$
\(\begin{gathered}
\text { Remaining }_2=\text { Remaining }_1-\text { Toy Store Expense } \\
\text { Remaining }_2=\frac{2}{5} A-\frac{2}{15} A
\end{gathered}\)
$$


To subtract, find a common denominator (15):

$$
\(\text { Remaining }_2=\frac{6}{15} A-\frac{2}{15} A=\frac{4}{15} A\)
$$

5. Candy Store Expense and Final Calculation

The problem states that he spent his last $\(\$ 0.80\)$ at the candy store. This means the money left after the toy store (Remaining $\({ }_2\)$ ) must equal $\(\$ 0.80\)$.

$$
\(\begin{aligned}
& \text { Remaining }_2=\$ 0.80 \\
& \qquad \frac{4}{15} A=0.80
\end{aligned}\)
$$


Solve for $A$ :

$$
\(\begin{gathered}
4 A=15 \times 0.80 \\
4 A=12.00 \\
A=\frac{12.00}{4} \\
A=3.00
\end{gathered}\)
$$


John's weekly allowance is $\(\$ 3.00\)$.

This matches option B.