To determine the longest delay for an $\backslash ($ n $\backslash$times n $\backslash )$ array multiplier, let's analyze the structure of such a multiplier. An array multiplier typically consists of a grid of AND gates for generating partial products and a network of adders $($often full adders$)$ to sum these partial products.

In an $\backslash ($ n $\backslash$times n $\backslash )$ array multiplier:

$1.$ The partial products are generated by $\backslash ($ n$^2\backslash )$ AND gates.

$2.$ These partial products are then summed up by a series of adders arranged in a grid.

The longest delay path is typically from the top left to the bottom right of the array. This path involves passing through a combination of adders and potentially AND gates.

For an $\backslash ($ n $\backslash$times n $\backslash )$ array multiplier:

$-$ There are $\backslash ($ n $\backslash )$ stages of addition $($with $\backslash ($ n$-1\backslash )$ adders in each column$)$ where each stage adds one more bit.

$-$ Each adder stage adds a delay of $\backslash ($ T$\_\{$ad$\}\backslash ).$

Considering the longest path through the array:

$-$ The delay through the AND gates for generating partial products is included but typically does not add to the overall path delay after the initial generation.

$-$ The longest path will traverse approximately $\backslash ((2$n $-1)\backslash )$ adders $($since we are moving diagonally across the array$),$ each contributing $\backslash ($ T$\_\{$ad$\}\backslash ).$

Thus, the total delay can be approximated as:

$\backslash [(2$n $-1)$ T$\_\{$ad$\}\backslash ]$

Given the options:

A$.3$nTad $+2$Tg

B$.$ On$($Tad $+$ Tg$)$

C$.(3$n$-4)$Tad $+$ Tg

D$.$ None of the above

The delay is not exactly captured by any of the options provided. Therefore, based on the typical understanding of array multipliers, the closest but not exactly precise answer would be:

D$.$ None of the above