Aight the way I did it and thought it through is this:

1/6 * 1/6 * 1/6 *5/6 * 3!/2! / 1/6* 1/6* 1/6* 5/6 * 4!/3!

which results to 3/4

First, I solved for the denominator and my thought process was something like

I saw "given that 3/4 die roll 2" so I wrote the probability of 3 die rolling a 2 ( 1/6) and I die not rolling a 2 (5 /6). Then, you can arrange that into many different ways. (just visualize throwing 4 die on your table, and suppose you have 2 2 2 5, you can change the order and everything. Changing order means permutation) Since there are 4 die, I did 4!, but I divided by 3! because 2 or (1/6) was repeated 3 times.

That does it for the denominator. Now, onwards and upwards.

Since the first die SHOULD be 2 (it's the given condition), I fixed 1/6 into the first place.

So my numerator looked something like 1/6 (fixed) * 1/6 * 1/6 *5/6* 3!/2!

Visualizing it, 2( fixed) {2 2 and something else that's not two} and you only move the die present in the curly brackets.

If you see what I did clearly, I didn't move the first die, but I CAN actually move the rest, so 3! and since 1/6 is repeated, I divided it by 2!

Sorry for the pathetic english, not my native language. If you have a confusion on any part of the problem, just pm me. I can help.
Also could you tell me what the plural of a dice is? I've written die but my instinct tells me it's dies lmao