Explanation:

1. Define original dimensions and area:

   Let $L\_0$ be the original length and $W\_0$ be the original width of the banner.

   The original area $A\_0 = L\_0 \times W\_0 = 2300$ square inches.

2. Determine the new dimensions:

   The length and width are each increased by $20\%$ .

   New length $L\_1 = L\_0 + 20\%$ of $L\_0 = L\_0 + 0.20L\_0 = 1.20L\_0$.

   New width $W\_1 = W\_0 + 20\%$ of $W\_0 = W\_0 + 0.20W\_0 = 1.20W\_0$.

3. Calculate the area of the copy:

   The area of the copy $A\_1 = L\_1 \times W\_1$ .

   Substitute the expressions for $L\_1$ and $W\_1$:

   $A\_1 = (1.20L\_0) \times (1.20W\_0)$

   $A\_1 = (1.20 \times 1.20) \times (L\_0W\_0)$

   $A\_1 = 1.44 \times A\_0$

4. Substitute the original area into the equation for the new area:

   $A\_1 = 1.44 \times 2300$

5. Calculate the result:

   $A\_1 = 1.44 \times 2300 = 3312$ square inches.

   (Calculation: $1.44 \times 2300 = \dfrac{144}{100} \times 2300 = 144 \times 23 = 3312$ )

6. Compare with the given options:

   @[A] $2,320$

   @[B] $2,340$

   @[C] $2,760$

   @[D] $3,312$ - This matches our calculation.

7. Conclusion:

   The area of the copy is $3,312$ square inches. The correct answer is D.