Answer: A

Solution:

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$

There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$

Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Each block $(5j+1)(5j+2)(5j+3)(5j+4)$ is $24\equiv-1\pmod{25},$ and there are $18$ blocks, so $A\equiv1\pmod{25}.$

After removing the $21$ factors of $5$ from $B,$ the remaining factors can be grouped as $$\begin{aligned} \dfrac{B}{5^{21}}={}&(1\cdot2\cdot3\cdot4) \\ &\cdot(6\cdot7\cdot8\cdot9) \\ &\cdot(11\cdot12\cdot13\cdot14) \\ &\cdot(16\cdot17\cdot18)(1\cdot2\cdot3) \\ &\equiv-1\pmod{25}. \end{aligned}$$

Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25}$ and the inverse of $2$ modulo $25$ is $13,$ we get $N\equiv-13\equiv12\pmod{25}.$

The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$

Thus, A is the correct answer.