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The correct answer is $518.4$. It's given that the area of the original poster is $360$ square inches. Let $l$ represent the length, in inches, of the original poster, and let $w$ represent the width, in inches, of the original poster. Since the area of a rectangle is equal to its length times its width, it follows that $360=lw$. It's also given that a copy of the poster is made in which the length and width of the original poster are each increased by $20\%$. It follows that the length of the copy is the length of the original poster plus $20\%$ of the length of the original poster, which is equivalent to $l+\frac{20}{100}l$ inches. This length can be rewritten as $l+0.2l$ inches, or $1.2l$ inches. Similarly, the width of the copy is the width of the original poster plus $20\%$ of the width of the original poster, which is equivalent to $w+\frac{20}{100}w$ inches. This width can be rewritten as $w+0.2w$ inches, or $1.2w$ inches. Since the area of a rectangle is equal to its length times its width, it follows that the area, in square inches, of the copy is equal to $\left(1.2l\right)\left(1.2w\right)$, which can be rewritten as $\left(1.2\right)\left(1.2\right)\left(lw\right)$. Since $360=lw$, the area, in square inches, of the copy can be found by substituting $360$ for $lw$ in the expression $\left(1.2\right)\left(1.2\right)\left(lw\right)$, which yields $\left(1.2\right)\left(1.2\right)\left(360\right)$, or $518.4$. Therefore, the area of the copy, in square inches, is $518.4$.